16: Acids and Bases

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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

O^2-

 

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

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

Answer

The conjugate base is SO₄^2-

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 (K_a = 6.8 * 10^-4)
2. HClO (K_a = 3.0 * 10^-8)
3. HNO2 (K_a = 4.5 *10^-4)
4. HCN (K_a = 4.9 * 10^-10)
5. Acetic Acid (K_a = 1.8 *10^-5)

Answer

d. HCN (K_a = 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  K_a=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?  K_a = 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?  K_a = 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 100K_a 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?  K_a=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 K_a 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 K_a 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 K_a 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 K_a 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 K_a 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 (K_a =3.0*10-⁸)
2. 0.01M HNO₂ (K_a = 4.5*10^-4)
3. 0.01M HF (K_a = 6.8*10^-4)
4. 0.01M HClO (K_a = 3.0*10^-8)
5. These will all exhibit the same percentage ionization

Answer

c. 0.01M HF (K_a = 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? (K_a = 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	CO3^2^-(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?  K_b(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? K_a(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?  K_a1(H₂CO₃)=4.5x10^-7, K_a2(H₂CO₃)=4.7x10^-11,  

Answer

bicarbonate is amphiprotic, and K_b' 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 K_b' (10^-8) is greater than K_a2 (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? K_a(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?    K_a1(H₂CO₃)=4.5x10^-7, K_a2(H₂CO₃)=4.7x10^-11,  

Answer

bicarbonate is amphiprotic, and K_b' 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 K_b of the conjugate base (K_b')

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

 

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

 

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

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 CO₃^2- 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)     +	CO3^2^-(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 SO₃^2- 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

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

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

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

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

Questions \(\PageIndex{5.7.f}\) through  \(\PageIndex{5.7.l}\) deal with the concentrations of species that exist in aqueous solutions of selenous acid. 

 

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

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is  [H₃O⁺] of H₂SeO₃?

Answer

The reaction is 

\[H_2SeO_3 \leftrightharpoons H^+ + HSeO_3^-  \;\;K_{a1}=2.7x10^{-5} \\ \; \\  HSeO_3^- \leftrightharpoons H^+ + SeO_3^{-2} \;\; K_{a2} = 2.5x10^{-9} \]

Since Ka₁ >1000Ka₂ you can consider all the hydronium to come from the first ionization, and that the second is insignificant.  This is logical, as in the first ionization you are pulling a proton from something neutral, while in the second you are pulling it from something negative, which just does not want to happen.

\[[H_3O]^+=\sqrt{K_{a1}[HA_i}=\sqrt{2.7*10^{-5}\left ( 0.50M \right )}=0.0037M\] 

 

 

 

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

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is pH of H₂SeO₃?

Answer

The reaction is 

\[H_2SeO_3 \leftrightharpoons H^+ + HSeO_3^-  \;\;K_{a1}=2.7x10^{-5} \\ \; \\  HSeO_3^- \leftrightharpoons H^+ + SeO_3^{-2} \;\; K_{a2} = 2.5x10^{-9} \]

This is actually the same question as \(\PageIndex{5.7.f}\), it is just asking you to express the hydronium ion concentration in terms of pH. Since Ka₁ >1000Ka₂ you can consider all the hydronium to come from the first ionization, and that the second is insignificant.  This is logical, as in the first ionization you are pulling a proton from something neutral, while in the second you are pulling it from something negative, which just does not want to happen.

\[pH=-log\left [ H^{+} \right ]=-log\sqrt{K_{a1}[HA_i]}=-log\sqrt{2.7*10^{-5}\left ( 0.50M \right )}=2.43 \nonumber \]

 

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

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is the concentration of HSeO₃^-?

Answer

The reaction is 

\[H_2SeO_3 \leftrightharpoons H^+ + HSeO_3^-  \;\;K_{a1}=2.7x10^{-5} \\ \; \\  HSeO_3^- \leftrightharpoons H^+ + SeO_3^{-2} \;\; K_{a2} = 2.5x10^{-9} \] 

Since Ka₁ >1000Ka₂ you can consider the second ionization as negligible and so [HSeO_3^-]=[H^+]

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

 

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

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is the concentration of H₂SeO₃?

Answer

The reaction is 

\[H_2SeO_3 \leftrightharpoons H^+ + HSeO_3^-  \;\;K_{a1}=2.7x10^{-5} \\ \; \\  HSeO_3^- \leftrightharpoons H^+ + SeO_3^{-2} \;\; K_{a2} = 2.5x10^{-9} \]

Since the amount dissociated it [H⁺]=[HSO₃^-] = 0.0037M, you simply subtract if from the initial concentration.

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

 

Note, this is a very weak acid and so it does not dissociate. 

 

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

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is the concentration of OH^-?

Answer

The reaction is 

\[H_2SeO_3 \leftrightharpoons H^+ + HSeO_3^-  \;\;K_{a1}=2.7x10^{-5} \\ \; \\  HSeO_3^- \leftrightharpoons H^+ + SeO_3^{-2} \;\; K_{a2} = 2.5x10^{-9} \]

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

 

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

Consider 0.50M of H₂SeO₃ for the following questions.

K_a1=2.7x10^-5                                               K_a2=2.5x10^-9

What is the concentration of SeO₃^2-?

Answer

 

Look at K₂ 

\[ K_{a2} = \frac{[H^+]}{[SeO_3^{-2}][HeSeO_3^-]}\]

Since \([H^+]=[HeSeO_3^-]\) the above equation becomes

\[ K_{a2} = \frac{\cancel{[H^+]}[SeO_3^{-2}]}{\cancel{[HeSeO_3^-]}}\]

\[K_{a2}=[SeO_3^{-2}]=2.5*10^{-9}M \nonumber \]

 

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)     +	HPO4^2^-(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	HPO4^2^-(aq)	⇌	H^+(aq)     +	PO4^3^-(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, these are all based on the structure of acetic acid, where you have one or more halogen relacing the non-titratable proton(s).  Chlorine is the most electronegative.  In the section on oxy acids with homologous structures we showed that the more electronegative chlorine pulled electron density towards it, and thus  made the oxygen-proton bond more polar, and thus more acidic.  The same thing is going on here. The trichloroacetic acid has three protons pulling electron density towards it, making it more acidic.  This is actually a nearest neighbor effect. 

 

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: AsH_3, 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 25^oC and 0.1atm?  (The solubility of CO₂ in pure water at 25^oC 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 CO₃^2- 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)     +	CO3^2^-(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 25^oC 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, [e-mail]. 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