16: Acids and Bases

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Chapter 16: Acids and Bases pdf

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

NaHSO₃is 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

NaHSO₃is 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?

(CH₃)₃NH⁺
CH₃COOH
HF
HNO₂
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 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.

What is the [H⁺]?
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

a weak base
a weak acid
a strong acid
a strong base
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?

pH = 3.21
pH = 12.59
pH = 7.93
pH = 9.82
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

water autoionizes very slowly
water autoionizes very quickly
water autoionizes only to a very small extent
the autoionization of water is exothermic
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?

18 M
100 M
55 M
0.100 M
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

HF (K_a = 6.8 * 10^-4)
HClO (K_a = 3.0 * 10^-8)
HNO2 (K_a = 4.5 *10^-4)
HCN (K_a = 4.9 * 10^-10)
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	K_a
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?

Cl^-
Br^-
NO₃^-
F^-
ClO₄^-

Answer: d. F^-

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

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

OH^-
Cl^-
NO₃^-
ClO^-
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?

H₂CO₃
H₂SO₄
HNO₃
HClO₄
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.

What is the pH of 0.001M Ba(OH)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?

0.01M HC₂H₂C₂ (K_a =3.0*10-⁸)
0.01M HNO₂ (K_a = 4.5*10^-4)
0.01M HF (K_a = 6.8*10^-4)
0.01M HClO (K_a = 3.0*10^-8)
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	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

\[pH+pOH=14 \nonumber \]

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

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

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 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.