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16: Acids and Bases

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    209176
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    Downloadable files

    16.1: Brønsted-Lowry Concept of Acids and Bases

    Textbook: Section 16.1

    Exercise \(\PageIndex{1.a}\)

    In the reaction, identify the acid and base.

    \(NH_{3}(g)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq)+OH^{-}(aq)\)

    Answer

    In this reaction H2O is donating a proton so it is the acid

    \(NH_{3}\) is accepting the proton so it is the base

     

    Exercise \(\PageIndex{1.b}\)

    What is the conjugate acid of \(NH_{3}\)?

    Answer

    A conjugate acid is a compound formed when an acid donates a proton to a base. So basically it is a base with a hydrogen ion added to it

    \(NH_{3}(g)+H^{+}\rightleftharpoons NH_{4}^{+}(aq)\nonumber \)

    So the conjugate acid of \(NH_{3}\) is \(NH_{4}^{+}\)

     

    Exercise \(\PageIndex{1.c}\)

    What is the conjugate base of H2O?

    Answer

    A conjugate base is what is left over after an acid has donated a proton. 

    \( H_{2}O \rightleftharpoons H^{+} + OH^{-} \nonumber \)

    So the Conjugate base of H2O is \(OH^{-}\)

     

    Exercise \(\PageIndex{1.d}\)

    What is the conjugate acid of NaHSO3?

    Answer

    NaHSO3 is an amphoteric substance, which means it can be an acid or a base

    \(HSO_{3}^{-}+H^{+}\rightarrow H_{2}SO_{3} \)

    So the conjugate acid of \(NaHSO_{3}\) is \(H_{2}SO_{3}\)

     

    Exercise \(\PageIndex{1.e}\)

    What is the conjugate base of NaHSO3?

    Answer

    NaHSO3 is an amphoteric substance, which means it can be an acid or a base

    \(HSO_{3}^{-}+OH^{-}\rightarrow SO_{3}^{-2}+H_{2}O\nonumber\)

    So the conjugate base of \(NaHSO_{3}\) is \(SO_{3}^{-2}\)

     

     

    Exercise \(\PageIndex{1.f}\)

    It is known that the hydride ion H- is a stronger base than OH-, what is(are) the product(s) of the reaction: H-(aq) +H2O(l) 

    Answer

    \(H^{-}(aq)+H_{2}O(l)\rightarrow H_{2}(g)+OH^{-}(aq)\nonumber\)

    base        acid        C.A.        C.B.

    Since H- is a stronger base than OH-, the reaction is reasonable to take place.

     

    Exercise \(\PageIndex{1.h}\)

    Which of the following is a Brønsted-Lowry acid?

    1. (CH3)3NH+
    2. CH3COOH
    3. HF
    4. HNO2
    5. all of these
    Answer

    e. all of these

     

    Exercise \(\PageIndex{1.i}\)

    What is the conjugate acid of NH3?

    Answer

    NH4+

     

    Exercise \(\PageIndex{1.j}\)

    What is the conjugate base of OH-?

    Answer

    O2-

     

    Exercise \(\PageIndex{1.k}\)

    What is the conjugate base of HSO4-? Conjugate acid?

    Answer

    The conjugate base is SO42-

    The conjugate acid is H2SO4

     

    16.2: Water and the pH Scale

    Textbook: Section 16.2

    Exercise \(\PageIndex{2.a}\)

    In a sample of lemon juice, [H+] is 6.2x10-4 M.  What is the pH?

    Answer

    \[pH=-log[H^{+}]=-log[6.2*10^{-4}M]=3.2\nonumber \]

     

     

    Exercise \(\PageIndex{2.b}\)

    A sample of detergent has a pH of 8.20.

    1. What is the [H+]?
    2. What is the [OH-]?
    Answer a.

    \[[H^{+}]=10^{-pH}=10^{-8.20}=6.3*10^-9M\nonumber \]

    Answer b.

    \[[OH^{-}]=\frac{K_{w}}{[H^{+}]}=\frac{10^{-14}}{6.3*10^{-9}}=1.6*10^{-6}\nonumber \]

     

    Exercise \(\PageIndex{2.c}\)

    What is the pH of an aqueous solution at 25°C in which [H+] is 0.0025 M?

    Answer

    \[pH=-log\left [ H^{+} \right ]=-log\left ( 0.0025M \right )=2.60\nonumber\]

     

    Exercise \(\PageIndex{2.d}\)

    What is the pH of a solution that contains 3.98 * 10-9 M hydronium ion at 25°C?

    Answer

    \[pH=-log\left [ H^{+} \right ]=-log\left ( 3.98*10^{-9}M \right )=8.400\nonumber\]

     

    Exercise \(\PageIndex{2.e}\)

    A 0.0035M aqueous solution of a compound has a pH=2.46. The compound is

    1. a weak base
    2. a weak acid
    3. a strong acid
    4. a strong base
    5. a salt
    Answer

    c. a strong acid

     

    Exercise \(\PageIndex{2.f}\)

    Calculate the pOH of a solution at 25°C that contains 1.94 * 10-10M hydronium ions.

    Answer

    \[pH=-log\left [ H^{+} \right ]=-log\left ( 1.94*10^{-10}M \right )=9.712 \nonumber\]

    \[pOH=14-pH=14-9.712=4.288\nonumber\]

     

    Exercise \(\PageIndex{2.g}\)

    Which solution below has the highest concentration of hydroxide ions?

    1. pH = 3.21
    2. pH = 12.59
    3. pH = 7.93
    4. pH = 9.82
    5. pH = 7.00
    Answer

    pH = 12.59

    \([OH^{-}]=10^{-12.59}=2.570*10^{-13}M \nonumber\)

     

    Exercise \(\PageIndex{2.h}\)

    What is the [H+] of an aqueous solution whose pH is 8.11?

    Answer

    \([H^{+}]=10^{-8.11}=7.76*10^{-9}M \nonumber\)

     

    Exercise \(\PageIndex{2.i}\)

    What is the concentration of H+ in a solution at 25°C with a pH of 7.35?

    Answer

    \([H^{+}]=10^{-7.35}=4.47*10^{-8}M \nonumber\)

     

    Exercise \(\PageIndex{2.j}\)

    The magnitude of Kw indicated that

    1. water autoionizes very slowly
    2. water autoionizes very quickly
    3. water autoionizes only to a very small extent
    4. the autoionization of water is exothermic
    5. the autoionization of water is endothermic
    Answer

    c. water autoionizes only to a very small extent

     

    Exercise \(\PageIndex{2.k}\)

    What is the concentration of water in pure water?

    1. 18 M
    2. 100 M
    3. 55 M
    4. 0.100 M
    5. 83 M
    Answer

    c. 55 M

     

    Exercise \(\PageIndex{2.l}\)

    What is the pH of 10-9M HCl (a very dilute strong acid)?

    Answer

    pH = 7, because an acid can not have a pH above 7. It is so dilute that the pH is is a consequence of the auto dissociation of the water.

     

    Exercise \(\PageIndex{2.m}\)

    What is the pH of 12.0M HCl (super strong acid)?

    Answer

    pH = -1.079

     

    Exercise \(\PageIndex{2.n}\)

    What is the pH of 10-9M NaOH (very dilute strong base)?

    Answer

    pH = 7,  because an basevcan not have a pH below 7. It is so dilute that the pH is is a consequence of the auto dissociation of the water.

    Exercise \(\PageIndex{2.o}\)

    What is the pH of 6.0M NaOH (very concentrated strong base)?

    Answer

    pH = 14.78

     

    Exercise \(\PageIndex{2.p}\)

    What is the pH of 10-6M HCl (dilute strong acid)?

    Answer

    \[pH=-log(10^{-6})=6\nonumber\]

     

    Exercise \(\PageIndex{2.q}\)

    What is the pH of 6.0M CH3COOH (concentrated acetic acid)?

    Answer

    \[pH=-log\sqrt{K_a[HA]_i} = -log(\sqrt{(1.8*10^{-5})*6})=1.98\nonumber\]

     

    16.3: Equilibrium Constants for Acids and Bases

    Textbook: Section 16.3

    Exercise \(\PageIndex{3.a}\)

    The following acids are listed in order of increasing strength. List their conjugate bases in order of increasing strength. HCN, CH3COOH, HF, HClO4:

    Answer

    ClO4-< F-< CH3COO-< CN-

     

    Exercise \(\PageIndex{3.b}\)

    Which one of the following is the weakest acid

    1. HF (Ka = 6.8 * 10-4)
    2. HClO (Ka = 3.0 * 10-8)
    3. HNO2 (Ka = 4.5 *10-4)
    4. HCN (Ka = 4.9 * 10-10)
    5. Acetic Acid (Ka = 1.8 *10-5)
    Answer

    d. HCN (Ka = 4.9 * 10-10)

     

    Exercise \(\PageIndex{3.c}\)

    Using the table below, which is the strongest acid?

    Acid Ka
    HOAc 1.8 * 10-5
    HCHO2 1.8 * 10-4
    HClO 3.0 * 10-8
    HF 6.8 * 10-4
    Answer

    HF with a  Ka=6.8 * 10-4

     

    Exercise \(\PageIndex{3.d}\)

    Which species form the following list would be the strongest Bronsted-Lowry base?

    1. Cl-
    2. Br-
    3. NO3-
    4. F-
    5. ClO4-
    Answer

    d. F-

     

    Exercise \(\PageIndex{3.e}\)

    Which of the following ions will act as a weak base in water?

    1. OH-
    2. Cl-
    3. NO3-
    4. ClO-
    5. None of these will act as a weak base in water
    Answer

    d. ClO-

     

    16.4: Acid-Base Properties of Salts

    Textbook: Section 16.4

    Exercise \(\PageIndex{4.a}\)

    Given that the Ka for gallic acid, (HC7H5O5) is 4.57*10-3, what is the Kb for the gallate ion (C7H5O5-)? T = 25°C

    Answer

    \[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{4.57*10^{-3}}=2.19*10^{-12} \nonumber \]

     

    Exercise \(\PageIndex{4.b}\)

    Kb for C5H5N is 1.4*10-9. Ka for C5H5NH+ is _____. T = 25°C

    Answer

    \[K_{a}=\frac{K_{w}}{K_{b}}=\frac{10^{-14}}{1.4*10^{-9}}=7.1*10^{-6} \nonumber \]

     

    Exercise \(\PageIndex{4.c}\)

    Ka for HF is 7.0*10-4. Kb for the fluoride ion is _____.

    Answer

    \[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{7.0*10^{-4}}=1.4*10^{-11} \nonumber \]

     

    16.5: Acid, Base & Salt Equilibrium Calculations

    Textbook: Section 16.5

    Strong Acids

    Exercise \(\PageIndex{5.1.a}\)

    What molar concentration of aqueous hydrochloric acid would have a pH = 9.50?

    Answer

    It is not possible for a solution of hydrochloric acid to have a pH = 9.50

     

    Exercise \(\PageIndex{5.1.b}\)

    What is the [H+] and pH of a 0.0037 M HBr solution at 25°C?

    Answer

    \[\left [ HBr \right ]=\left [ H^{+} \right ]=0.0037M\nonumber\]

    \[pH=-log\left [ H^{+} \right ]=-log\left ( 0.0037M \right )=2.43\nonumber\]

     

    Exercise \(\PageIndex{5.1.c}\)

    Which of the following acids is not a strong acid?

    1. H2CO3
    2. H2SO4
    3. HNO3
    4. HClO4
    5. HCl
    Answer

    a. H2CO3

     

    Exercise \(\PageIndex{5.1.d}\)

    What is the pH of 0.055M of HCl?

    Answer

    \[H^{+}=0.055M\nonumber\]

    \[pH=-log[H^{+}]=-log[0.055M]=1.26\nonumber \]

     

    Exercise \(\PageIndex{5.1.e}\)

    What is the pH of an aqueous solution in which the molar concentration of HCl is 1.3 * 10-11?

    Answer

    You would think

    \[pH = -log\left [ H^{+} \right ]=-log\left ( 1.3*10^{-11} \right )=10.89\nonumber\]

    But it is 7, and the pH is dictated by the water

     

    Exercise \(\PageIndex{5.1.f}\)

    The pH of a 0.030 M HCl solution at 25°C is _____.

    Answer

    \[pH=-log\left [ H^{+} \right ]=-log\left ( 0.030M \right )=1.52\nonumber\]

     

    Strong Bases

    Exercise \(\PageIndex{5.2.a}\)

    What molar concentration of aqueous barium hydroxide would have pH = 12.25?

    Answer

    \[pOH=14-pH=14-12.25=1.75\nonumber\]

    \[\left [ OH^{-} \right ]=10^{-1.75}=0.01778M\nonumber\]

    \[\left [ Ba\left ( OH \right )_{2} \right ]=\frac{\left [ OH^{-} \right ]}{2}=\frac{0.01778M}{2}=0.008891M\nonumber\]

     

     

    Exercise \(\PageIndex{5.2.b}\)

    Consider the strong diprotic base Ba(OH)2.

    1. What is the pH of 0.001M  Ba(OH)2?
    2. What is the p0H of 0.001M Ba(OH)2?
    Answer a.

    \[[OH^{-}]=0.001*2=0.002M\nonumber \]

    \[[H^{+}]=\frac{K_{w}}{[OH^{-}]}=\frac{10^{-14}}{2.0*10^{-3}}=5.0*10^{-12}M\nonumber \]

    \[pH=-log[H^{+}]=-log[5.0*10^{-12}M]=11.30\nonumber \]

    Answer b.

    \[[OH^{-}]=0.001*2=0.002M\nonumber \]

    \[pOH=-log[OH^{-}]=-log[2.0*10^{-3}M]=2.70\nonumber \]

     

    Exercise \(\PageIndex{5.2b}\)

    If 0.56g of CaO is dissolved in the water to make 1.0L solution.  What is the pH of the solution?

    Answer

    \[CaO(aq)\rightleftharpoons Ca^{+2}(aq) +2O{-2}(aq)\nonumber \]

    \[O^{-2}(aqs)+H_{2}O(l)\rightleftharpoons Ca^{+2}(aq) +2OH-(aq)  \\ \; \\ so \\ \; \\ CaO(aq)+H_{2}O(l)\rightleftharpoons Ca^{+2}(aq) +2OH-(aq)\nonumber ]\]

     

    \[\frac{\frac{0.56g CaO}{\frac{56gCaO}{mol}}*\frac{2\,mol\,OH^-}{1\,mol\,CaO}}{1.0L}=0.02\,M\,OH^-\nonumber \]

    \[pOH=-log[OH^{-}]=-log[2.0*10^{-2}M]=1.70\nonumber \]

    \[pH=14-pOH=14-1.70=12.3\nonumber \]

     

    Exercise \(\PageIndex{5.2.c}\)

    What is the pH of a 0.015 M solution of barium hydroxide?

    Answer

    \[Ba(OH)_{2}\rightarrow Ba^{2+} + 2OH^{-}\nonumber\]

    \[[OH^{-}]=2*M=2*0.015M=0.03M\nonumber\]

    \[pOH=-log[OH^{-}]=-log(0.03M)=1.52\nonumber\]

    \[pH=14-pOH=14-1.52=12.48\nonumber\]

     

    Exercise \(\PageIndex{5.2.d}\)

    What is the [OH-] and pH of a 0.035M KOH solution at 25°C?

    Answer

    \[\left [KOH\right ]=\left [OH^{-}\right ]=0.035M\nonumber\]

    \[pH=14-pOH =14-(-log [OH^{-}])\nonumber\]

    \[14-1.46=12.54\nonumber\]

     

    Exercise \(\PageIndex{5.2.e}\)

    The pH of a 0.011 M NaOH solution at 25°C is _____.

    Answer

    \[pOH=-log\left (OH^{-}\right )=-log\left (0.011M\right )=1.96\nonumber\]

    \[pH=14-pOH=14-1.96=12.04\nonumber\]

     

    Exercise \(\PageIndex{5.2.f}\)

    What is the pH of a 0.053 M solution of Ca(OH)2?

    Answer

    \[\left [ OH^{-} \right ]=2*M=2*0.053M=0.11M\nonumber\]

    \[pOH=-log\left [ OH^{-} \right ]=-log\left ( 0.11M \right )=0.96\nonumber\]

    \[pH=14-pOH=14-0.96=13.04\nonumber\]

     

    Weak Acids

    Exercise \(\PageIndex{5.3.a}\)

    What is the pH of 0.20M aqueous HF?  Ka = 6.8x10-4

    Answer

    \[HF(aq)\rightleftharpoons H^{+}(aq)+F^{-}(aq) \nonumber \]

    \[[HF]_{i}>100K_{a} \nonumber \]

    Therefore,

    \[[H^{+}]=\sqrt{K_{a}[HF]_{i}}=\sqrt{(6.8*10^{-4})*0.20}=1.2*10^{-2}M \nonumber \]

    \[pH=-log[H^{+}]=-log[1.2*10^{-2}M]=1.93 \nonumber \]

     

    Exercise \(\PageIndex{5.3.b}\)

    What is the pH of 0.0050M aqueous HF?  Ka = 6.8x10-4

    Answer
    R HF(aq) H+(aq)          + F-(aq)
    I 0.0050   0 0
    C -x   +x +x
    E 0.0050-x   x x

    \(K_{a}=\frac{x^{2}}{0.0050-x}\) Note 100Ka is not less than [HA]i so we need to use the quadratic formula.

    \[K_{a}\left ( 0.0050-x \right )=x^{2}\Rightarrow x^{2}+K_{a}x-K_{a}\left ( 0.0050 \right ) \nonumber \]

    \[x^{2}+0.00068x-3.4*10^{-5} \nonumber \]

    \[x=\frac{-0.00068\pm \sqrt{\left ( 0.00068 \right )^{2}-4(1)(-3.4*10^{-6})}}{2} \nonumber \]

    \[x=\frac{-0.00068\pm 0.00375}{2} \nonumber \]

    \[x=[H^{+}]=0.001535M \nonumber \]

    \[pH=-log[H^{+}]=-log[0.001535M]=2.8 \nonumber \]

     

    Exercise \(\PageIndex{5.3.c}\)

    Which one is more acidic, 0.2MHF or 0.02MHCl?

    Answer

    \[0.02HCl:\,pH=-log[0.02]=1.70 \nonumber \]

    \[0.2HF:\,pH=-log(\sqrt{K_{a}[HA]_{i}})=-log(\sqrt{7.2*10^{-4}[0.2]})=1.93 \nonumber \]

    The more dilute strong acid HCl is more acidic than the more concentrated weak acid HF.

     

    Exercise \(\PageIndex{5.3.d}\)

    What is the pH of 0.04M of NH4Cl?  Ka=5.6x10-10

    Answer

    \[NH_{4}^{+}(aq)\rightleftharpoons NH_{3}(aq)+H^{+}(aq)\,\,\,K_{a}=5.6*10^{-10} \nonumber \]

    \[\left [NH_{4}^{+}\right ]_{i}>100K_{a}\nonumber\]

    Therefore,

    \[[H^{+}]=\sqrt{K_{a}[NH_{4}^{+}]_{i}}=\sqrt{5.6*10^{-10}*(0.04)}=4.73*10^{-6}M \nonumber \]

    \[pH=-log[H^{+}]=-log[4.73*10^{-6}M]=5.32 \nonumber \]

     

    Exercise \(\PageIndex{5.3.e}\)

    A 0.10M solution of lactic acid (HC3H5O3, one acidic hydrogen) has a pH of 2.45.  What is the Ka for lactic acid?

    Answer

    \[\left [ H^{+} \right ]=10^{-2.45}=0.0035M \nonumber \]

    R HC3H5O3(aq) C3H5O3-(aq)          + H+(aq)
    I 0.10M   0 0
    C -x   +x +x
    E 0.10-x   x x

    \[K_{a}=\frac{x^{2}}{0.10-x}=\frac{0.0035^{2}}{0.10-0.0035}=1.30*10^{-4} \nonumber \]

    16.5.3

    Exercise \(\PageIndex{5.3.f}\)

    A 0.15M aqueous solution of the weak acid HA at 25°C has a pH of 5.35. What is the value of Ka for HA?

    Answer
    R HA H+     + A-
    I 0.15M   0 0
    C -x   +x +x
    E 0.15-x   x x

    \[Ka=\frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=\frac{(x)(x)}{(0.15-x)} \nonumber \]

    \[\left [ H^{+} \right ]=x=10^{-pH}=10^{-5.35}=4.47*10^{-6}M \nonumber \]

    \[Ka=\frac{(x)(x)}{(0.15-x)}=\frac{(4.47*10^{-6})(4.47*10^{-6})}{(0.15-(4.47*10^{-6}))}=\frac{2.00*10^{-11}}{0.150}=1.33*10^{-10} \nonumber \]

    Exercise \(\PageIndex{5.3.g}\)

    The Ka of HClO is 3.0*10-8. What is the pH at 25°C of an aqueous solution that is 0.020M in HClO?

    Answer

    \[[HClO]_{i}>100K_{a} \nonumber \]

    Therefore,

    \[[H^{+}]=\sqrt{K_{a}[HClO]_{i}}=\sqrt{(3.0*10^{-8})*0.20}=2.5*10^{-5}M \nonumber \]

    \[pH=-log[H^{+}]=-log[2.5*10^{-5}M]=4.61 \nonumber \]

    Or if you really want to spend the time you can do the RICE diagram.....

    R HClO H+     + ClO-
    I 0.020M   0 0
    C -x   +x +x
    E 0.020-x   x x

    \[Ka=\frac{\left [ H^{+} \right ]\left [ ClO^{-} \right ]}{\left [ HClO \right ]}=\frac{(x)(x)}{(0.020-x)} \nonumber \]

    \[3.0*10^{-8}=\frac{(x)(x)}{(0.020-x)} \nonumber \]

    \[\left [ H^{+} \right ]=x=2.5*10^{-5}M \nonumber \]

    \[pH=-log\left [ H^{+} \right ]=-log\left ( 2.5*10^{-5} \right )=4.61 \nonumber \]

     

    Exercise \(\PageIndex{5.3.h}\)

    The Ka of HF is 6.8*10-4. What is the pH of a 0.35M solution of HF? 

    Answer

    \[[HF]_{i}>100K_{a} \nonumber \]

    Therefore,

    \[[H^{+}]=\sqrt{K_{a}[HF]_{i}}=\sqrt{(6.8*10^{-4})*0.35}=0.0154M \nonumber \]

    \[pH=-log[H^{+}]=-log[0.1054M]=1.81 \nonumber \]

    Or if you really want to spend the time you can do the RICE diagram.....

    R HF H+     + F-
    I 0.35M   0 0
    C -x   +x +x
    E 0.35-x   x x

    \[Ka=\frac{\left [ H^{+} \right ]\left [ F^{-} \right ]}{\left [ H \right ]}=\frac{(x)(x)}{(0.35-x)} \nonumber \]

    \[6.8*10^{-4}=\frac{(x)(x)}{(0.35-x)} \nonumber \]

    \[\left [ H^{+} \right ]=x=1.5*10^{-2}M \nonumber \]

    \[pH=-log\left [ H^{+} \right ]=-log\left ( 1.5*10^{-2} \right )=1.81 \nonumber \]

    Exercise \(\PageIndex{5.3.i}\)

    A 0.25M solution of the weak acid HA has a pH of 4.15. What is the value of Ka for HA? 

    Answer

    \[HA\rightleftharpoons H^{+}+A^{-} \nonumber \]

    \[\left [ H^{+} \right ]=\left [ A^{-} \right ]=10^{-pH}=10^{-4.15}=7.08*10^{-5}M \nonumber \]

    Note how x<<[HA]i and so in the next step we can ignore the extent of reaction (amount dissociated)

    \[Ka=\frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]-x} \approxeq \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]} \nonumber \]

     

    \[Ka=\frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=\frac{\left ( 7.08*10^{-5}M \right )\left ( 7.08*10^{-5}M \right )}{\left ( 0.25M \right )}=2.0*10^{-8} \nonumber \]

     

    Exercise \(\PageIndex{5.3.j}\)

    In which of the following aqueous solution does the weak acid exhibit the highest percentage ionization?

    1. 0.01M HC2H2C2 (Ka =3.0*10-8)
    2. 0.01M HNO2 (Ka = 4.5*10-4)
    3. 0.01M HF (Ka = 6.8*10-4)
    4. 0.01M HClO (Ka = 3.0*10-8)
    5. These will all exhibit the same percentage ionization
    Answer

    c. 0.01M HF (Ka = 6.8*10-4)

     

    Percent Ionization

    Exercise \(\PageIndex{5.4.a}\)

    What is the percent ionization of 0.25 aqueous HF?  Ka = 6.8x10-4

    Answer

    \[HF(aq)\rightleftharpoons H^{+}(aq)+F^{-}(aq)\nonumber \]

    \[\left [HF\right ]_{i}>100Ka\nonumber\]

    Therefore,

    \[\left [H^{+}\right ]=\sqrt{Ka\left [HF\right ]_{i}}=\sqrt{6.8*10^{-4}*0.25}=1.3*10^{-2}M\nonumber\]

    \[\%\text{Ionized}=\frac{\text{amount dissociated}}{\text{Initial concentration}}(100)=\frac{1.3*10^{-2}}{0.25}=5.2\%\nonumber\]

     

    Exercise \(\PageIndex{5.4.b}\)

    What is the percent ionization of 0.0055 aqueous HF?  Ka = 6.8x10-4

    Answer
    R HF(aq) H+(aq)     + F-(aq)
    I 0.0055M   0 0
    C -x   +x +x
    E 0.0055-x   x x

    \[Ka=\frac{x^{2}}{0.0055-x}=6.8*10^{-4}\nonumber\]

    \[x=\left [H^{+}\right ]=1.60*10^{-3}M\nonumber\]

    \[\frac{1.60*10^{-3}}{0.0055}=29.5\%\nonumber\]

     

    Exercise \(\PageIndex{5.4.c}\)

    What is the percent ionization of 0.05M of NH4Cl?  Ka=5.6x10-10

    Answer

    \(NH_{4}^{+}(aq)\rightleftharpoons NH_{3}(aq)+H^{+}(aq)\), \(Ka=5.6*10^{-10}\)

    \[\left [NH_{4}^{+}\right ]_{i}>100Ka\nonumber\]

    \[\left [H^{+}\right ]=\sqrt{Ka\left [NH_{4}^{+}\right ]_{i}}=\sqrt{5.6*10^{-11}*0.005}=5.29*10^{-6}M\nonumber\]

    \[\%\text{Ionized}=\frac{\text{amount dissociated}}{\text{Initial concentration}}(100)=\frac{5.29*10^{-6}}{0.05}*100\%=0.011\%\nonumber\]

     

    Exercise \(\PageIndex{5.4.d}\)

    What is the percent ionization of a 0.002M solution of H2CO3? Ka1=4.3x10-7, Ka2 = 5.6x10-11

    Answer

    \[Ka_{1}=4.3*10^{-7}, Ka_{2}=5.6*10^{-11}\nonumber\]

    \[K_{a1}>1000 k_{a2} \;\;\; so \;\; \left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=\sqrt{Ka_{1}*0.002}=2.9*10^{-5}M\nonumber\]

    \[\%\text{Ionized}=\frac{\text{amount dissociated}}{\text{Initial concentration}}(100)=\frac{2.9*10^{-5}}{0.002}*100\%=1.45\%\nonumber\]

     

    Exercise \(\PageIndex{5.4.e}\)

    What is the percent ionization of 0.04M of hydrazoic acid (HN3)?  Ka=1.9x10-5

    Answer

    \[HN_{3}(aq)\rightleftharpoons H^{+}(aq)+N_{3}^{-}(aq)\nonumber\]

    \[\left [HN_{3}\right ]_{i}>100Ka\nonumber\]

    Therefore,

    \[\left [H^{+}\right ]=\sqrt{Ka\left [HN_{3}\right ]_{i}}=\sqrt{1.9*10^{-5}*0.04}=8.7*10^{-4}M\]

    \[\%\text{Ionized}=\frac{\text{amount dissociated}}{\text{Initial concentration}}(100)=\frac{8.7*10^{-4}}{0.04}=2.2\%\nonumber\]

     

    Exercise \(\PageIndex{5.4.f}\)

    What is the percent ionization of hypochlorous acid (HClO) in a 0.015 M aqueous solution of HClO at 25C? (Ka = 3.0 * 10-8)

    Answer

    \[HClO(aq)\rightleftharpoons H^{+}(aq)+ClO^{-}(aq)\nonumber\]

    \[\left [HClO\right ]_{i}>100Ka\nonumber\]

    Therefore,

    \[\left [H^{+}\right ]=\sqrt{Ka\left [HClO\right ]_{i}}=\sqrt{3.0*10^{-8}*0.015}=2.12*10^{-5}M\]

    \[\%\text{Ionized}=\frac{\text{amount dissociated}}{\text{Initial concentration}}(100)=\frac{2.2*10^{-5}}{0.015}(100)=0.14 \% \nonumber\]

     

     

    Weak Base

    Exercise \(\PageIndex{5.5.a}\)

    What is the concentration of OH- in 0.10M of ethylamine (C2H5NH2)? Kb=6.4x10-4

    Answer

    \[\left [ C_{2}H_{5}NH_{2} \right ]>100K_{b} \nonumber \]

    \[\left [ OH^{-} \right ]=\sqrt{\left (6.4*10^{-4}\right )*0.10}=0.008M \nonumber \]

     

    Exercise \(\PageIndex{5.5.b}\)

    What is the concentration of OH- in 0.005M of ethylamine (C2H5NH2)? Kb=6.4x10-4

    Answer
    R C2H5NH2(aq)     + H2O(l) C2H5NH3+(aq)     + OH-(aq)
    I 0.005M     0 0
    C -x     +x +x
    E 0.005-x     x x

    \[K_{b}=\frac{x^{2}}{0.005-x}=6.4*10^{-4} \nonumber \]

    \[x=1.5*10^{-3} \nonumber \]

     

    Exercise \(\PageIndex{5.5.c}\)

    Which solution is more basic, 0.10M of ethylamine or 0.01M of NaOH?

    Note, you solved 0.01M ethylamine two problems ago

    Answer

    NaOH, 0.01M of NaOH has a pOH of 2.00

    Exercise \(\PageIndex{5.5.d}\)

    A solution of NH3 has a pH of 10.25.  What is the concentration of the solution? Kb=1.8x10-5

    Answer

    \[[H^+]=10^{-10.25}=5.6*10^{-11}   \;\; K_w=[H^+][OH^-] \; \\ \; [OH^-]=\frac{K_w}{[H^+]}=\frac{10^{14}}{5.6x10^{-11}}=1.8x10^{-4} \nonumber \]

     

    \[\begin{align}K_{b} & =\frac{x^{2}}{[NH_3]_i-x } \nonumber \\ \; \\ K_b\left(  [NH_3]_i -x \right) & = x^2  \nonumber \\ \; \\  K_b [NH_3]_i -K_bx & = x^2 \nonumber \\ \; \\ K_b [NH_3]_i  &= x^2 + K_bx \nonumber \\ \; \\ [NH_3]_i &=\frac{x^2 + K_bx}{K_b}  \nonumber \\ \; \\ [NH_3]_i &=\frac{\left( 1.8x10^{-4} \right)^2+1.8x10^{-5}}{1.8x10^{-5}} \nonumber \\ \; \\ [NH_3]_i &=2.0*10^{-3}M  \end{align} \]

    If you want, you can look at the RICE diagram

    R NH3(aq)     + H2O(l) NH4+(aq) OH-(aq)
    I [NH3]i     0 0
    C -x     +x +x
    E [NH3]i-x     x x

     

     

    Exercise \(\PageIndex{5.5.e}\)

    If 1.06g of Na2CO3 is dissolved in plenty of water to make 1.0L of the solution, what is the pH of the solution? Kb=1.8x10-4

    Answer

    \[\frac{\frac{1.06g}{106g/mol}}{1.0L}=0.01M \nonumber \]

    \[[OH^-]= \sqrt{K_b[B]}=\sqrt{\left( 1.8x10^{-4} \right) 0.01}=0.0013 \;\; so \;\;p[OH]=-log(0.0013)=2.87 \;\;and\;\;pH=14-pOH=11.13

    Or if you insist on using the RICD diagram, you can....

     

    R CO32-(aq)     + H2O(l) HCO3-(aq)     + OH-(aq)
    I 0.01M     0 0
    C -x     +x +x
    E 0.01-x     x x

     

    \[K_{b}=\frac{x^{2}}{0.01-x}=1.8*10^{-4} \nonumber \]

    \[x=1.31464*10^{-3}M \nonumber \]

    \[pOH=2.87 \nonumber \]

    \[pH=14.00-2.87=11.13 \nonumber \]

     

    Exercise \(\PageIndex{5.f}\)

    The pH of a 0.10M solution of a weak base is 9.82. What is the Kb for this base?

    Answer

    \[pH+pOH=14 \nonumber \]

    \[pOH=14-9.82=4.18 \nonumber \]

    \[ [OH^{-}] = 10^{-pOH} = 10^{-4.18}= 6.61*10^{-5} M \nonumber \]

    d

    R B     + H2O(l) B+         + OH-(aq)
    I 0.10M     0 0
    C -x     +x +x
    E 0.10-x     x x

    \[OH^-]=[B^+]=x ;and\; K_b=\frac{x^2}{[B]_i-x} \approxeq \frac{x^2}{[B]_i} \nonumber\]

    This is the same eq as \([OH^-]=\sqrt{K_b[B]_i}\)

    \[K_{b}=\frac{\left ( 6.61*10^{-5} \right )^{2}}{0.1}=4.37*10^{-8} \nonumber \]

     

    pH of Various Salts

    Exercise \(\PageIndex{5.6.a}\)

    What is the pH of 0.05M of NH4Cl?  Kb(NH3) = 1.8x10-5

    Answer

    \[NH_{4}^{+}(aq)\rightleftharpoons NH_{3}(aq)+H^{+}(aq) \nonumber \]

    \[(K_a'(NH_4^+) = \frac{K_w}{K_b}=\frac{10^{-14}}{(1.8x10^{-5}}=5.6*10^{-10}\nonumber \]

    \[\left [NH_{4}^{+}\right ]_{i}>100Ka'\nonumber \]

    Therefore,

    \[\left [H^{+}\right ]=\sqrt{Ka'\left [NH_{4}^{+}\right ]_{i}}=\sqrt{5.6*10^{-10}*0.05}=5.3*10^{-6}M\nonumber \]

    \[pH=-log\left (5.3*10^{-6}M\right )=5.28\nonumber \]

    To solve in one equation

    \[pH=-log\sqrt{\frac{K_w}{K_b}\left[B^+\right]} = 14-(-log\sqrt{\frac{10^{-14}}{1.8x10^{-5}}\left[0.05\right]} ) =5.3 \nonumber \]

    NOTE: the answer has one sig fig because 0.05 has one sig fig, by doing it in one step you truncate at the end and this gives the best answer.

     

    Exercise \(\PageIndex{5.6.b}\)

    What is the pH of 0.05M of NaClO? Ka(HClO)=4.0x10-8.

    Answer

    hypochlorite is the conjugate base of hypochlorous acid and the salt forms hypochlorous acid

    \[ClO^{-}(aq)+H_{2}O(l)\rightleftharpoons HClO(aq)+OH^{-}(aq) \nonumber\]

    \[(K_b'(ClO^-) = \frac{K_w}{K_b'}=\frac{10^{-14}}{4.0x10^{-8}}=2.5*10^{-7}\nonumber \]

    \[\left [ClO^{-}\right ]_{i}>100K_{b}\nonumber \]

    \[\left [OH^{-}\right ]=\sqrt{K_{b}\left [ClO^{-}\right ]_{i}}=\sqrt{2.5*10^{-7}*0.05}=1.118*10^{-4}M\nonumber \]

    \[pOH=-log\left (1.3*10^{-4}M\right )=3.95\nonumber \]

    \[pH=14.00-3.89=10.0\nonumber \]

    To solve in one equation

    \[pH=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14 +log\sqrt{\frac{10^{-14}}{4.0x10^{-8}}\left[0.05\right]}  =10.0 \nonumber \]

    NOTE: the answer has one sig fig because 0.05 has one sig fig, by doing it in one step you truncate at the end and this gives the best answer.

     

    Exercise \(\PageIndex{5.6.c}\)

    What is the pH of 0.05M of NaHCO3?  Ka1(H2CO3)=4.5x10-7, Ka2(H2CO3)=4.7x10-11,  

    Answer

    bicarbonate is amphiprotic, and Kb' comes from the first ionization constant

    \[HCO_{3}^{-}(aq)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq)+OH^{-}(aq) \nonumber \]

    \[(K_b'(HCO_3^-) = \frac{K_w}{K_a}=\frac{10^{-14}}{4.5x10^{-7}}=2.22*10^{-8}\nonumber \]

     

    Since the Kb' (10-8) is greater than Ka2 (10-11) it is a stronger base than an acid, so: 

    \[\left [OH^{-}\right ]=\sqrt{K_{b}\left [HCO_{3}^{-}\right ]_{i}}=\sqrt{2.22*10^{-8}*0.05}=3.33*10^{-5}M\nonumber \]

    \[pOH=-log\left (3.4*10^{-5}M\right )=4.48\nonumber \]

    \[pH=14.00-4.48=9.5\nonumber \]

    To solve in one equation

    \[pH=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14 + (og\sqrt{\frac{10^{-14}}{4.5x10^{-7}}\left[0.05\right]} =9.5 \nonumber \]

    NOTE: the answer has one sig fig because 0.05 has one sig fig, by doing it in one step you truncate at the end and this gives the best answer.

     

    Exercise \(\PageIndex{5.6.d}\)

    What is the pH of 0.05M of KF? Ka(HF)=6.3x10-4.

    Answer

    fluoride is the conjugate base of hydrofluoric acid

    \[F^{-}(aq)+H_{2}O(l)\rightleftharpoons HF(aq)+OH^{-}(aq)\nonumber \]

    \[(K_b'(F^-) = \frac{K_w}{K_a}=\frac{10^{-14}}{6.3x10^{-4}}=1.58710^{-11}\nonumber \]

    \[\left [F^{-}\right ]_{i}>100K_{b}'\nonumber \]

    Therefore,

    \[\left [OH^{-}\right ]=\sqrt{K_{b}\left [F^{-}\right ]_{i}}=\sqrt{1.587*10^{-11}*0.05}=8.909*10^{-7}M\nonumber \]

    \[pOH=-log\left (8.909*10^{-7}M\right )=6.05\nonumber \]

    \[pH=14.00-6.06=7.95\nonumber \]

    To solve in one equation

    \[pH=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14+log\sqrt{\frac{10^{-14}}{6.3x10^{-4}}\left[0.05\right]} =7.95\nonumber \]

    Exercise \(\PageIndex{5.6.e}\)

    How many grams of NaHCO3 will be used to make a 1.0L solution that has a pH = 9.0?    Ka1(H2CO3)=4.5x10-7, Ka2(H2CO3)=4.7x10-11,  

    Answer

    bicarbonate is amphiprotic, and Kb' comes from the first ionization constant

    \[HCO_{3}^{-}(aq)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq)+OH^{-}(aq) \nonumber \]

    \[(K_b'(HCO_3^-) = \frac{K_w}{K_a}=\frac{10^{-14}}{4.5x10^{-7}}=2.22*10^{-8}\nonumber \]

    R HCO3-(aq)     + H2O(l) H2CO3(aq)     + OH-(aq)
    I [HCO3-(aq)]i     0 0
    C -x     +x +x
    E [HCO3-(aq)]i-x     x x

    \[pH+pOH=14 \;\;so\;\;\; pOH=14-pH=14-9=5 \;\;\;[OH^-] = 10^{-pOH}=10^{-5}] \nonumber \]

     

    \[K_{b}=\frac{x^{2}}{[HCO_3^-]-x} \approxeq \frac{x^2}{[HCO_3^-]} \nonumber \]

    \[[HCO_3^-]=\frac{x^2}{K_b}=\frac{[10^{-5}]^2}{2.22x10^{-8}} = 0.0045M \nonumber \]

    \[\frac{0.0045mol \;NaHCO_3}{l}\left(\frac{84.007g \; NaHCO_3}{mpl}\right)=0.378g=0.38g \nonumber \]

     

    Exercise \(\PageIndex{5.6.f}\)

    Ka of HA is 7.5*10-9. What is the pH of a 0.15M solution of NaA?

    Answer

    A- is the conjugate base of HA

    \[A^-(aq) + H_2O(l)\rightleftharpoons HA(aq)+OH^{1}(aq) \nonumber \]

    \[(K_b'(A^-) = \frac{K_w}{K_a}=\frac{10^{-14}}{(7.5x10^{-9}}=1.333*10^{-6}\nonumber \]

    \[\left [A^-\right ]_{i}>100Ka'\nonumber \]

    \[\left [OH^{-}\right ]=\sqrt{K_{b}\left [A^{-}\right ]_{i}}=\sqrt{1.33x10^{-6}*0.15}=4.472*10^{-4}M\nonumber \]

    \[pOH=-log\left (4.472*10^{-4}M\right )=3.35\nonumber \]

    \[pH=14.00-3.76=10.65\nonumber \]

    To solve in one equation

    \[pH=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14 +log\sqrt{\frac{10^{-14}}{7.5x10^{-9}}\left[0.15\right]}  =10.65 \nonumber \]

    NOTE: the answer has two sig figs, by doing it in one step you truncate at the end and this gives the best answer.

    Exercise \(\PageIndex{5.6.g}\)

    Calculate the pOH of a 0.0827M aqueous sodium cyanide solution at 25°C (for CN-, Kb = 4.9*10-10).

    Answer
    \[pOH=-log[OH-]=-log\sqrt{K_b[CN^-]}=-log\sqrt{4.9x10^{-10}(0.827)}=5.20\nonumber \]

     

    Exercise \(\PageIndex{5.6.h}\)

    Determine the pH of a 0.15M solution of KF. For hydrofluoric acid, Ka = 7.0*10-4.

    Answer

    First we will do this in one equation, and then do it stepwise, noting we use primes (') to indicate K's of conjugates.

    \[pH=14-pOH=14-\sqrt{K_b'[A^-]}=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14 +log\sqrt{\frac{10^{-14}}{7.0x10^{-5}}\left[0.15\right]}  =8.16 \nonumber \]

    Now we will do it stepwise

    \[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{7.0*10^{-4}}=1.4*10^{-11} \nonumber \]

    \[K_{b}=\frac{\left [ HF \right ]\left [ OH^{-} \right ]}{\left [ F^{-} \right ]}=\frac{(x)(x)}{0.15-x} \approxeq\frac{(x)(x)}{0.15}1.4*10^{-11} \nonumber \]

    \[x=\left [ OH^{-} \right ]=1.45*10^{-6}M \nonumber \]

    \[pOH=-log\left [ OH^{-} \right ]=-log\left [ 1.45*10^{-6}M \right ]=5.84 \nonumber \]

    \[pH=14-pOH=14-5.84=8.16 \nonumber \]

    Exercise \(\PageIndex{5.6.i}\)

    Calculate the pH of 0.726M anilinium hydrochloride, (C6H5NH3Cl) solution in water given that Kb for aniline is 3.83*10-4.

    Answer

    Anilinium is the conjugate acid of aniline (C6H5N2)

    \[C_6H_5NH_4^+ \rightleftharpoons  C_6H_5NH_2 + H^+\]

    In one equation:

    \[pH=-log\sqrt{K_a'[C_6H_5NH_4^+]}=-log\sqrt{\frac{K_w}{K_b}[C_6H_5NH_4^+]}=-log\sqrt{\frac{1x10^{-14}}{3.83x10^{-4}}[0.726]}=5.36 \nonumber \]

    Taking this out stepwise....

    \[K_{a}'=\frac{K_{w}}{K_{b}}=\frac{10^{-14}}{3.83*10^{-4}}=2.61*10^{-11} \nonumber \]

    \[K_{a}'=\frac{\left [ C_{6}H_{5}NH_{2} \right ]\left [ H^{+} \right ]}{\left [ C_{6}H_{5}NH_{3}^{+} \right ]}=\frac{(x)(x)}{0.726-x}\approxeq \frac{(x)(x)}{0.726}=2.61*10^{-11} \nonumber \]

    solving for x

    \[x=\left [ H^{+} \right ]=4.35*10^{-6} M  \\  pH=-log\left [ H^{+} \right ]=-log\left [ 4.35*10^{-6}M \right ]=5.36 \nonumber \]

     

    Exercise \(\PageIndex{5.6.j}\)

    The Ka for formic acid (HCHO2) is 1.8*10-4. What is the pH of a 0.35M solution of sodium formate (NaCHO2)?

    Answer

    Firstst we will do this in one equation, and then do it stepwise, noting we use primes (') to indicate K's of conjugates.

    \[pH=14-pOH=14-\sqrt{K_b'[A^-]}=14-(-log\sqrt{\frac{K_w}{K_a}\left[A^-\right]} )= 14 +log\sqrt{\frac{10^{-14}}{1.8x10^{-4}}\left[0.35\right]}  =8.64 \nonumber \]

    Now we will do it stepwise. First calculate Kb of the conjugate base (Kb')

    \[K_{b}'=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{1.8*10^{-4}}=5.55*10^{-11} \nonumber \]

     

    \[[OH^-]=\sqrt{K_b'[A^-]}=\sqrt{5.55x10^{-11}[0.35]}=4.41x10^{-6} \nonumber \]

     

    \[pOH=-log[OH^-]=-log4.41x10^{-6}=5.36 \]

    pH=14-pOH=14-5.36= 8.64 \nonumber \]

     

     

    Polyprotic Acids and Bases

     

    Exercise \(\PageIndex{5.7.a}\)

    What is the pH of a 0.002M solution of H2CO3? Ka1=4.3x10-7, Ka2 = 5.6x10-11

    Answer

    \[Ka_{1}=4.3*10^{-7},\,Ka_{2}=5.6*10^{-11} \nonumber \]

    \[\frac{Ka_{1}}{Ka_{2}}>1000 \nonumber \]

    \[\left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=\sqrt{Ka_{1}*0.002}=2.9*10^{-5}M \nonumber \]

    \[pH=4.53 \nonumber \]

     

    Exercise \(\PageIndex{5.7.b}\)

    What is the concentration of CO32- ion in the solution in Q16.3.11?

    Answer

    \[\left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=2.9*10^{-5}M \nonumber \]

    R HCO3-(aq) H+(aq)     + CO32-(aq)
    I 2.9*10-5   2.9*10-5 0
    C -x   -x +x
    E 2.9*10-5-x   2.9*10-5-x x

    \[Ka_{2}=\frac{\left (2.9*10^{-5}-x\right )x}{2.9*10^{-5}-x}=5.6*10^{-11} \nonumber \]

    \[\left [CO_{3}^{2-}\right ]=5.6*10^{-11}M \nonumber \]

     

    Exercise \(\PageIndex{5.7.c}\)

    What is the pH of a 0.05M of sulfurous acid (H2SO3)? Ka1=1.7x10-2, Ka2=6.4x10-8

    Answer

    \[\frac{Ka_{1}}{Ka_{2}}>1000 \nonumber \]

    R H2SO3(aq) H+(aq) HSO3-(aq)
    I 0.05   0 0
    C -x   +x +x
    E 0.05-x   x x

    \[Ka_{1}=\frac{x^{2}}{0.05-x}=1.7*10^{-2} \nonumber \]

    \[x=2.2*10^{-2}M \nonumber \]

    \[pH=1.66 \nonumber \]

     

    Exercise \(\PageIndex{5.7.d}\)

    What is the concentration of SO32- ion in the solution in Q16.3.13?

    Answer

    \[\left [H^{+}\right ]=\left [HSO_{3}^{-}\right ]=2.2*10^{-2}M \nonumber \]

    \[Ka_{2}=\frac{\left (2.2*10^{-2}-x\right )x}{2.2*10^{-2}-x}=6.4*10^{-8} \nonumber \]

    \[\left [SO_{3}^{2-}\right ]=6.4*10^{-8}M \nonumber \]

     

    Exercise \(\PageIndex{5.7.e}\)

    What is the concentration of a sample solution of H2CO3 that has a pH = 4.50?

    Answer

    \[10^{-4.50}=3.16*10^{-5} \nonumber \]

    \[\left [H^{+}\right ]=\sqrt{Ka_{1}*\left [H_{2}CO_{3}\right ]}=3.16*10^{-5}M \nonumber \]

    \[Ka_{1}=4.3*10^{-7} \nonumber \]

    \[\left [H_{2}CO_{3}\right ]=0.0023M \nonumber \]

     

    Exercise \(\PageIndex{5.7.f}\)

    Consider 0.50M of H2SeO3 for the following questions.

    Ka1=2.7x10-5                                               Ka2=2.5x10-9

    What is the pH of H2SeO3?

    Answer

    \[pH=-log\left [ H^{+} \right ]=-log\left [ HSeO_{3}^{-} \right ]=-log\sqrt{2.7*10^{-5}\left ( 0.50M \right )}=2.43 \nonumber \]

     

    Exercise \(\PageIndex{5.7.g}\)

    Consider 0.50M of H2SeO3 for the following questions.

    Ka1=2.7x10-5                                               Ka2=2.5x10-9

    What is the concentration of H+?

    Answer

    \[pH=-log\left [ H^{+} \right ]=\left [ HSeO_{3}^{-} \right ]=\sqrt{2.7*10^{-5}\left ( 0.50M \right )}=0.0037M \nonumber \]

     

    Exercise \(\PageIndex{5.7.h}\)

    Consider 0.50M of H2SeO3 for the following questions.

    Ka1=2.7x10-5                                               Ka2=2.5x10-9

    What is the concentration of HSeO3-?

    Answer

    \[pH=-log\left [ H^{+} \right ]=\left [ HSeO_{3}^{-} \right ]=\sqrt{2.7*10^{-5}\left ( 0.50M \right )}=0.0037M \nonumber \]

     

    Exercise \(\PageIndex{5.7.j}\)

    Consider 0.50M of H2SeO3 for the following questions.

    Ka1=2.7x10-5                                               Ka2=2.5x10-9

    What is the concentration of H2SeO3?

    Answer

    \[H_{2}SeO_{3}=0.50M-\left [ H^{+} \right ]=0.50M-0.0037M=0.4963M=0.50M \nonumber \]

     

    Exercise \(\PageIndex{5.7.k}\)

    Consider 0.50M of H2SeO3 for the following questions.

    Ka1=2.7x10-5                                               Ka2=2.5x10-9

    What is the concentration of OH-?

    Answer

    \[\left [ OH^{-} \right ]=\frac{K_{w}}{\left [ H^{+} \right ]}=\frac{10^{-14}}{\left [ H^{+} \right ]}=2.7*10^{-12}M \nonumber \]

     

    Exercise \(\PageIndex{5.7.l}\)

    Consider 0.50M of H2SeO3 for the following questions.

    Ka1=2.7x10-5                                               Ka2=2.5x10-9

    What is the concentration of SeO32-?

    Answer

    \[Ka_{2}=\frac{\left [ H^{+} \right ]\left [ CO_{3}^{2-} \right ]}{\left [ HCO_{3}^{-} \right ]}=\left [ CO_{3}^{2-} \right ]=2.5*10^{-9}M \nonumber \]

     

    Exercise \(\PageIndex{5.7.m}\)

    An aqueous solution of phosphoric acid has a concentration of 2.5M. (Ka1 = 7.5*10-3, Ka2 = 6.2*10-8, Ka3 = 4.2*10-13)

    1. What is the pH?
    2. What is the molar concentration of phosphate ion?
    Answer a.

    First dissociation of phosphoric acid:

    \[H_{3}PO_{4}(aq)\rightleftharpoons H^{+}(aq)+H_{2}PO_{4}^{-}(aq)\,\,\,K_{a1}=7.5*10^{-3} \nonumber \]

    \[H_{2}PO_{4}^{-}\rightleftharpoons H^{+}(aq)+HPO_{4}^{2-}(aq)\,\,\,K_{a2}=6.2*10^{-8} \nonumber \]

    \[HPO_{4}^{2-}(aq)\rightleftharpoons H^{+}(aq)+PO_{4}^{3-}(aq)\,\,\,K_{a3}=4.2*10^{-13} \nonumber \]

    R H3PO4(aq) H+(aq)     + H2PO4-(aq)
    I 2.5M   0 0
    C -x   +x +x
    E 2.5+x   x x

    \[K_{a1}=\frac{\left [H^{+}\right ]\left [H_{2}PO_{4}^{-}\right ]}{\left [H_{3}PO_{4}\right ]}=\frac{(x)(x)}{(2.5-x)}=7.5*10^{-3} \nonumber \]

    \[x=\left [H^{+}\right ]=0.13323M \nonumber \]

     

    Second dissociation of phosphoric acid:

    R H2PO4-(aq) H+(aq)     + HPO42-(aq)
    I 0.13323M   0.13323M 0
    C -y   +y +y
    E 0.13323-y   0.13323+y y

    \[K_{a2}=\frac{\left [ H^{+} \right ]\left [ HPO_{4}^{2-} \right ]}{\left [ H_{2}PO_{4}^{-} \right ]}=\frac{(y)(0.13323+y)}{0.13323-y}=6.2*10^{-8} \nonumber \]

    (Assume <<0.13323 M)

    \[\left [ H^{+} \right ]=K_{a2}=6.2*10^{-8} \nonumber \]

    The [H+] from the second step is negligible.

     

    Third dissociation of phosphoric acid:

    R HPO42-(aq) H+(aq)     + PO43-(aq)
    I 6.2*10-8M   0.13323M 0
    C -z   +z +z
    E 6.2*10-8-z   0.13323+z y

    \[K_{a3}=\frac{\left [ H^{+} \right ]\left [ PO_{4}^{3-} \right ]}{\left [ HPO_{4}^{2-} \right ]}=\frac{\left ( z \right )\left ( 0.13323+z \right )}{\left ( 6.2*10^{-8} \right )-z}=4.2*10^{-13} \nonumber \]

    \[z=1.4*10^{-19}M \nonumber \]

    The [H+] from the third step is negligible.

    \[pH=-log\left [ H^{+} \right ]=-log\left [ 0.13323 \right ]=0.88 \nonumber \]

    Answer b.

    First dissociation of phosphoric acid:

    R H3PO4(aq) H+(aq)     + H2PO4-(aq)
    I 2.5M   0 0
    C -x   +x +x
    E 2.5+x   x x

    \[K_{a1}=\frac{\left [H^{+}\right ]\left [H_{2}PO_{4}^{-}\right ]}{\left [H_{3}PO_{4}\right ]}=\frac{(x)(x)}{(2.5-x)}=7.5*10^{-3} \nonumber \]

    \[x=\left [H_{2}PO_{4}^{-}\right ]=0.13323M \nonumber \]

     

    For the second dissociation:

    \[K_{a2}=\frac{\left [ H^{+} \right ]\left [ HPO_{4}^{2-} \right ]}{\left [ H_{2}PO_{4}^{-} \right ]}=\frac{(x)(x)}{(0.13323-x)}=6.2*10^{-8} \nonumber \]

    \[x=\left [ HPO_{4}^{2-} \right ]=1.92*10^{-4}M \nonumber \]

     

    For the third dissociation:

    \[K_{a3}=\frac{\left [ H^{+} \right ]\left [ PO_{4}^{3-} \right ]}{\left [ HPO_{4}^{2-} \right ]}=\frac{(x)(x)}{(1.92*10^{-4})-x}=4.2*10^{-13} \nonumber \]

    \[x=\left [ PO_{4}^{3-} \right ]=1.92*10^{-19}M \nonumber \]

     

    Molecular Structure

    Exercise \(\PageIndex{6.a}\)

    Of the following, which is the strongest acid?

    1. HIO
    2. HIO4
    3. HIO2
    4. HIO3
    5. The acid strength of all these is nearly the same.
    Answer

    b. HIO4

    Exercise \(\PageIndex{6.b}\)

    Of the following, the acid strength of _____ is the greatest.

    1. CH3COOH
    2. ClCH2COOH
    3. Cl2CHCOOH
    4. Cl3CCOOH
    5. BrCH2COOH
    Answer

    d. Cl3CCOOH

     

    Exercise \(\PageIndex{6.c}\)

    Of the following, _____ is the strongest acid.

    1. Cl3C-COOH
    2. H3C-COOH
    3. Br2C-COOH
    4. F3C-COOH
    5. Br2ClC-COOH
    Answer

    d.F3C-COOH

     

    Exercise \(\PageIndex{6.d}\)

    Which of the following acids will be the strongest?

    1. H2SO4
    2. HSO4-
    3. H2SO3
    4. H2SeO4
    5. HSO3-
    Answer

    a.  H2SO4

     

    Exercise \(\PageIndex{6.e}\)

    The more electronegative X is, the _____ polar will be the H-X bond and the _____ easily the H-X bond is broken, making HX more _____ acidic.

    1. more, less, weakly
    2. more, more, weakly
    3. more, more, strongly
    4. more, less, strongly
    5. less, less, strongly
    Answer

    c. more, more, strongly

     

    Exercise \(\PageIndex{6.f}\)

    Which one is more acidic, HNO2 or HNO3?

    Answer

    HNO3

     

    Exercise \(\PageIndex{6.g}\)

    Which one is more acidic H3AsO3 or H3AsO4?

    Answer

    H3AsO4

     

    Exercise \(\PageIndex{6.h}\)

    List the acids in order of increasing acid strength: HClO2, HBrO2, HIO2

    Answer

    HIO2< HBrO2< HClO2

     

    Exercise \(\PageIndex{6.i}\)

    List the compounds in order of increasing acid strength: AsH3, HBr, NaH, H2O

    Answer

    NaH< AsH3< H2O< HBr

     

    Exercise \(\PageIndex{6.j}\)

    List the compounds in order of increasing acid strength: H2TeO3, H2TeO4, H2O

    Answer

    H2O< H2TeO3< H2TeO4

     

    Acid Anhydrides

    Exercise \(\PageIndex{6.1.a}\)

    What is the pH of the solution if 0.05mol of SO3 is dissolved in the water to make 1.0L solution?  (SO3 is very soluble.)

    Answer

    \[SO_{3}(g)+H_{2}O(l)\rightleftharpoons H_{2}SO_{4}(aq) \nonumber \]

    \[H_{2}SO_{4}(aq)\rightleftharpoons H^{+}(aq)+HSO_{4}^{-}(aq) \nonumber \]

    \[\frac{0.05mol}{1.0L}=0.05M  pH=-log\left (0.05M\right )=1.30 \nonumber \]

     

    Exercise \(\PageIndex{6.1.b}\)

    How many grams of SO3 is needed to make a 1.0L solution that has a pH=1.0?

    Answer

    \[SO_{3}(g)+H_{2}O(l)\rightleftharpoons H_{2}SO_{4}(aq) \nonumber \]

    \[H_{2}SO_{4}(aq)\rightleftharpoons H^{+}(aq)+HSO_{4}^{-}(aq) \nonumber \]

    \[\left [H^{+}\right ]=0.1M \nonumber \]

    \[0.1M*1.0L=0.1mol \nonumber \]

    \[0.1mol*80g/mol=8.0g \nonumber \]

     

    Exercise \(\PageIndex{6.1.c}\)

    What is the pH of the solution if 0.002mol of CO2 is dissolved in the water to make 1.0L solution at 25oC and 0.1atm?  (The solubility of CO2 in pure water at 25oC and 0.1atm is 0.0037M.) Ka1=4.3x10-7, Ka2 = 5.6x10-11

    Answer

    \[CO_{2}(g)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq) \nonumber \]

    \[H_{2}CO_{3}(aq)\rightleftharpoons H^{+}(aq)+HCO_{3}^{-}(aq) \nonumber \]

    \[\frac{0.002mol}{1.0L}=0.002M \nonumber \]

    \[\left [H^{+}\right ]=\sqrt{4.3*10^{-7}*0.002}=2.93*10^{-5}M\]

    \[pH=-log\left (2.93*10^{-5}M\right )=4.53 \nonumber \]

     

    Exercise \(\PageIndex{6.1.d}\)

    What is the concentration of CO32- ion in the solution in Q 16.5.3?

    Answer

    \[\left [H^{+}\right ]=\left [HCO_{3}^{-}\right ]=2.93*10^{-5}M \nonumber \]

    R HCO3-(aq) H+(aq)     + CO32-(aq)
    I 2.93*10-5   2.93*10-5 0
    C -x   +x +x
    E 2.93*10-5-x   2.93*10-5+x x

    \[Ka_{2}=\frac{\left (2.93*10^{-5}-x\right)x}{2.93*10^{-5}+x}=5.6*10^{-11} \nonumber \]

    \[\left [CO_{3}^{2-}\right]=5.6*10^{-11} \nonumber \]

     

    Exercise \(\PageIndex{6.1.e}\)

    How many grams of CO2 is needed at 25oC and 0.1atm to make a 1.0L solution that has a pH=4.50?

    Answer

    \[pH=4.50 \nonumber \]

    \[\left [H^{+}\right]=10^{-4.50}=3.16*10^{-5}M \nonumber \]

    \[CO_{2}(g)+H_{2}O(l)\rightleftharpoons H_{2}CO_{3}(aq) \nonumber \]

    R H2CO3(aq) H+(aq)     + HCO3-(aq)
    I M   0 0
    C -x   +x +x
    E M-x   3.16*10-5 x

    \[Ka=\frac{\left ( 3.16*10^{-5} \right )^{2}}{M-3.16*10^{-5}}=4.3*10^{-7} \nonumber \]

    \[M=2.35*10^{-3}mol/L \nonumber \]

    \[2.35M*1.0L=2.36mol \nonumber \]

    \[2.35mol*44g/mol=0.10g \nonumber \]

     

    Basic Anhydrides

    Exercise \(\PageIndex{6.2.a}\)

    What is the pH of the solution that is prepared by dissolving 0.62g of Na2O in enough water to make 1.0L?

    Answer

    \[Na_{2}O(s)+H_{2}O(l)\rightleftharpoons 2NaOH(aq) \nonumber \]

    \[NaOH(aq)\rightleftharpoons Na^{+}(aq)+OH^{-}(aq) \nonumber \]

    \[\left [NaOH\right ]=\left [OH^{-}\right ]=\frac{2*\frac{0.62g}{62g/mol}}{1.0L}=0.02M \nonumber \]

    \[pH=14-log\left [OH^{-}\right ]=14.0-\left (-log\left (0.02M\right )\right )=12.3 \nonumber \]

     

    Exercise \(\PageIndex{6.2.b}\)

    How many grams of Na2O is needed to make a 1.0L solution that has a pH=13.0?

    Answer

    \[pH=13.0 \nonumber \]

    \[\left [OH^{-}\right]=10^{-\left (14-13\right )}=0.10M \nonumber \]

    \[0.10M*1.0L=0.10mol \nonumber \]

    \[Na_{2}O(s)+H_{2}O(l)\rightleftharpoons 2NaOH(aq) \nonumber \]

    \[\frac{Na_{2}O}{2NaOH}=\frac{x mol}{0.10mol} \nonumber \]

    \[x=0.05mol \nonumber \]

    \[0.05mol*62g/mol=3.10g \nonumber \]

     

    Exercise \(\PageIndex{6.2.c}\)

    What is the pH of the solution that is prepared by dissolving 0.56g of CaO in the water to make 1.0L?

    Answer

    \[CaO(s)+H_{2}O(l)\rightleftharpoons Ca\left (OH\right )_{2}(aq) \nonumber \]

    \[Ca\left (OH\right )_{2}(aq)\rightleftharpoons Ca^{2+}(aq)+2OH^{-}(aq) \nonumber \]

    \[2\left [Ca\left (OH\right )_{2}\right]=\left [OH^{-}\right ]=\frac{2*\frac{0.56g}{56g/mol}}{1.0L}=0.02M \nonumber \]

    \[pH=14-\left (-log\left [ OH^{-} \right ]\right )=14.0-1.70=12.3 \nonumber \]

     

    Exercise \(\PageIndex{6.2.d}\)

    How many grams of CaO is needed to make a 1.0L solution that has a pH = 13.0?

    Answer

    \[pH=13.0 \nonumber \]

    \[\left [OH^{-}\right]=10^{-\left ( 14-13 \right )}=0.10M \nonumber \]

    \[0.10M*1.0L=0.10mol \nonumber \]

    \[CaO(s)+H_{2}O(l)\rightleftharpoons Ca\left (OH\right )_{2}(aq) \nonumber \]

    \[Ca\left (OH\right )_{2}(aq)\rightleftharpoons Ca^{2+}(aq)+2OH^{-}(aq) \nonumber \]

    \[\frac{2OH^{-}}{Ca\left (OH\right )_{2}}=\frac{Ca\left (OH\right )_{2}}{CaO}=\frac{0.10mol}{xmol} \nonumber \]

    \[x=0.05mol \nonumber \]

    \[0.05mol*56g/mol=2.80g \nonumber \]

     

    Exercise \(\PageIndex{6.2.e}\)

    If 100.0ml of the solution in Q 16.5.9 is transferred to a 500.0ml container, and plenty water was added to fill it up, what is the pH of the solution?

    Answer

    \[\left [OH^{-}\right]=10^{-\left ( 14-13 \right )}=0.10M \nonumber \]

    \[Ca\left (OH\right )_{2}(aq)\rightleftharpoons Ca^{2+}(aq)+2OH^{-}(aq) \nonumber \]

    \[\frac{\left [ Ca\left (OH\right )_{2} \right ]}{\left [ OH^{-} \right ]}=\frac{1}{2} \nonumber \]

    \[\left [ Ca\left (OH\right )_{2} \right ]=0.05M \nonumber \]

    \[\frac{0.05M*0.10L}{0.50L}=0.01M \left [ OH^{-} \right ]=0.02M \nonumber \]

    \[pH=14-\left ( -log\left [ OH^{-} \right ] \right )=14.0-1.7=12.3 \nonumber \]

     

    16.7: Lewis Concept of Acids and Bases

    Textbook: Section 16.7

    Exercise \(\PageIndex{7.a}\)

    CO can form complexes with metals, eg. Fe(CO)5, Ni(CO)4.  Is CO a lewis acid or a lewis base?

    Answer

    CO has a lone pair of electrons so it is a Lewis base

     

    Exercise \(\PageIndex{7.b}\)

    In Q 16.7.a, Is the metal, such as Fe(II), a lewis acid or a lewis base?

    Answer

    Fe(II) accepts the electron pair so it is a Lewis acid

     

    Exercise \(\PageIndex{7.c}\)

    In the reaction, \(Zn\left ( OH \right )_{2}(s)+2OH^{-}(aq)\rightleftharpoons Zn\left ( OH \right )_{4}^{2-}(aq)\), Which one is the lewis acid?

    Answer

    \(Zn\left ( OH \right )_{2}\) accepts an electron pair so it is a lewis acid

     

    Exercise \(\PageIndex{7.d}\)

    In the reaction, \(CO_{2}+O^{2-}\rightarrow CO_{3}^{2-}\), Which one is the lewis acid and lewis base?

    Answer

    Lewis acid: \(CO_{2}\)

    Lewis base: \(O^{2-}\)

     

    Exercise \(\PageIndex{7.e}\)

    Can CH3NH2 be a lewis acid or a lewis base?

    Answer

    Lewis base

     

    Exercise \(\PageIndex{7.f}\)

    Trimethylamine (CH3)3N can react with diborane B2H6  after its dissociation to form (CH3)3N-BH3.  Which one is the lewis acid?  Which one is the lewis base?

    Answer

    Lewis acid: B2H6

    Lewis base: (CH3)3N

     

    Exercise \(\PageIndex{7.g}\)

    In the reaction, \(H_{2}NOH(aq)+HCl(aq)\rightarrow \left [ H_{3}NOH \right ]Cl(aq)\), which one is the lewis acid?

    Answer

    \(H_{2}NOH\)

     

    Exercise \(\PageIndex{7.h}\)

    In the reaction, \(SO_{2}(g)+BF_{3}(g)\rightarrow O_{2}S-BF_{3}(s)\), which one is the lewis acid? and the lewis base?

    Answer

    Lewis acid: BF3

    Lewis base: SO2

     

    Exercise \(\PageIndex{7.i}\)

    Which one of the following cannot act as a Lewis base?

    1. Cl-
    2. NH3
    3. BF3
    4. CN-
    5. H2O
    Answer

    C. BF3

     

    General Questions

    Textbook: Section 16

    Exercise \(\PageIndex{8.a}\)

    A 0.1M aqueous solution of _____ will have the highest pH.

    1. KCN, Ka of HCN = 4.0*10-10
    2. NH2NO3, Kb of NH3 = 1.8*10-5
    3. NaOAc, Ka of HOAc = 1.8*10-5
    4. NaClO, Ka of HClO = 3.2*10-8
    5. NaHS, Kb of HS- = 1.8*10-7
    Answer

    a. KCN, Ka of HCN = 4.0*10-10

     

    Exercise \(\PageIndex{8.b}\)

    A 0.1M solution of _____ has a pH of 7.0.

    1. Na2S
    2. KF
    3. NaNO3
    4. NH3Cl
    5. NaF
    Answer

    c. NaNO3

     

    Exercise \(\PageIndex{8.c}\)

    Which of the following possesses the greatest concentration of hydroxide ion?

    1. a solution with a pH of 3.0
    2. a 1 * 10-4 M solution of HNO3
    3. a solution with a pOH of 12.0
    4. pure water
    5. a 1 * 10-3 M solution of NH4Cl
    Answer

    c. solution with pH of 12.0

     

    Textbook: 16: Acids and Bases

    Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

    • Liliane Poirot

    16: Acids and Bases is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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