5.13 Practice Problems

Attempt these problems as if they were real exam questions in an exam environment.

Only look up information if you get severely stuck. Never look at the solution until you have exhausted all efforts to solve the problem. Having to look up information or the solution should be an indicator that the previous layers (1–5) in the Structured Learning Approach have not been mastered.

What is the K_a for a 0.05 M monoprotic weak acid solution that has a pH of 3.85 (at 25 °C)?

4.0 × 10^–7
1.4 × 10^–4
2.3 × 10^–6
3.5 × 10²
3.85

Solution

Answer: A

Concept: pH and Equilibrium

We are given the pH of the solution. Therefore, we can get the hydronium ion concentration of the solution at equilibrium.

\[\begin{align*} [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 10^{-\mathrm{pH}} \\[1.5ex] &= 10^{-3.85} \\[1.5ex] &= 1.41\times 10^{-4}~M \end{align*}\]

We relate this the hydronium ion concentration of the weak acid solution. Consider the acid-ionization reaction for a weak, monoprotic acid.

\[\mathrm{HA}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{A^-}(aq)\]

The contribution of hydronium ions from water is much less than the hydronium ions present (i.e. 1 × 10^–7 << 1.41 × 10^–4) so we can safely ignore it.

Filling out an ICE table gives

	HA(aq)	H₃O⁺(aq)	A^–(aq)
I	0.05	≈0	0
C	-x	+x	+x
E	0.05-x	1.41 × 10^–4	x

We see that x = 1.41 × 10^–4. Therefore, we can obtain all the equilibrium concentrations directly.

\[\begin{align*} [\mathrm{HA}]_{\mathrm{eq}} &= 0.05 - x \\[1.5ex] &= 0.05 - 1.41\times 10^{-4} \\[1.5ex] &= 0.049859~M \\[2.5ex] [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 1.41 \times 10^{-4}~M \\[2.5ex] [\mathrm{A^-}]_{\mathrm{eq}} &= x \\[1.5ex] &= 1.41 \times 10^{-4}~M \end{align*}\]

Write out the equilibrium expression and solve for K_a.

\[\begin{align*} K_{\mathrm{a}} &= \dfrac{[\mathrm{H_3O}^+][\mathrm{A^-}]}{[\mathrm{HA}]}\\[1.5ex] &= \dfrac{(1.41 \times 10^{-4}~M)(1.41 \times 10^{-4}~M)}{0.049859~M}\\[1.5ex] &= 3.987\times 10^{-7} \end{align*}\]

What is the pH of water at 60 °C? K_w = 9.311 × 10^–14
1. 6.5
2. 6.9
3. 7.0
4. 7.4
5. 8.2
Solution

Answer: A

Concept: K_w

Remember that the pH of water changes with temperature because K_w changes! Recall the auto-ionization of water reaction

\[\mathrm{2H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH}^-(aq)\]

where the equilibrium expression is

\[K_{\mathrm{w}} = [\mathrm{H_3O^+}][\mathrm{OH^-}]\]

K_w is only equal to 1.0 × 10^–14 at 25 °C. In this problem, water is at 60 °C and the K_w is given.

We can convert K_w to pK_w

\[\begin{align*} \mathrm{p}K_{\mathrm{w}} &= -\log K_{\mathrm{w}} \\[1.5ex] &= -\log 9.311\times 10^{-14} \\[1.5ex] &= 13.031 \end{align*}\]

and then use

\[\mathrm{p}K_{\mathrm{w}} = \mathrm{pH} + \mathrm{pOH}\]

to get the pH. Since we’re considering pure water, we know that pH = pOH.

\[\begin{align*} \mathrm{p}K_{\mathrm{w}} &= \mathrm{pH} + \mathrm{pOH} \\[1.5ex] &= 2(\mathrm{pH}) \longrightarrow \\[1.5ex] \mathrm{pH} &= \dfrac{\mathrm{p}K_{\mathrm{w}}}{2} \\[1.5ex] &= \dfrac{13.031}{2} \\[1.5ex] &= 6.515 \end{align*}\]
What is the pOH of an aqueous solution containing 0.40 M HI (at 25 °C)?
1. 1.40
2. 3.24
3. 7.68
4. 8.43
5. 13.6
Solution

Answer: E

Concept: Strong Acid Equilibrium and pH

This is an strong acid equilibrium problem followed by a pH to pOH conversion.

Write the Balanced Reaction

HI is a strong acid so we know that K_a is very large.

\[\mathrm{HI}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{H_3O^+}(aq) + \mathrm{I^-}(aq) \qquad K_{\mathrm{a}} >> 1\]

This means that HI will cease to exist at equilibrium as all of it gets converted into products. Therefore,

\[[\mathrm{HI}]_{\mathrm{i}} \approx [\mathrm{H_3O^+}]_{\mathrm{eq}} = 0.40~M\]

Find pH

\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}]_{\mathrm{eq}} \\[1.5ex] &= -\log (0.40~M) \\[1.5ex] &= 0.3979 \end{align*}\]

Find pOH

Because we are at 25 °C, we know K_w = 1.0 × 10^–14 and, therefore, pK_w = 14.

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 - 0.3979 \\[1.5ex] &= 13.60 \end{align*}\]
Identify the Brønsted-Lowry acid in the following reaction.

\[\mathrm{NO_3}^-(aq) + \mathrm{H_3O^+}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{HNO_3}(aq)\]
1. H₂O
2. HNO₃
3. H₃O⁺
4. NO₃^–
5. none of these
Solution

Answer: C

Concept: Acid-Base Definitions

A Brønsted-Lowry acid is a proton donor. In the given reaction, hydronium donates a proton to NO₃^–.
A substance is found to be an electron-pair donor. What acid/base definition best describes this substance?
1. Arrhenius base
2. Lewis base
3. Brønsted-Lowry acid
4. Lewis acid
Solution

Answer: B

Concept: Acid-Base Definitions

An electron-pair donor is a proton donor is a Lewis base.
0.25 mol of a strong, monoprotic acid is dissolved in water resulting in a 500 mL aqueous solution. What is the pH (at 25 °C)?
1. 0.3
2. 0.6
3. 1.2
4. 2.4
5. 3.3
Solution

Answer: A

Concept: Strong Acid Equilibrium and pH

This is an strong acid equilibrium problem.

Write the Balanced Reaction

HI is a strong acid so we know that K_a is very large.

\[\mathrm{HA}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{H_3O^+}(aq) + \mathrm{A^-}(aq) \qquad K_{\mathrm{a}} >> 1\]

This means that HI will cease to exist at equilibrium as all of it gets converted into products.

Find Molar Concentration of Strong Acid

\[\begin{align*} [\mathrm{HA}] &= \dfrac{n}{V} \\[1.5ex] &= \dfrac{0.25~\mathrm{mol}}{500~\mathrm{mL} \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right )} \\[1.5ex] &= 0.5~M \end{align*}\]

Find Hydronium Ion Concentration at Equilibrum

\[[\mathrm{HA}]_{\mathrm{i}} \approx [\mathrm{H_3O^+}]_{\mathrm{eq}} = 0.5~M\]

Find pH

\[\begin{align*} \mathrm{pH} &= -\log [\mathrm{H_3O^+}]_{\mathrm{eq}} \\[1.5ex] &= -\log (0.50~M) \\[1.5ex] &= 0.30103 \end{align*}\]
The K_a for hydrofluoric acid is 7.2 × 10^–4 at 25 °C. What is the corresponding K_b for its conjugate base?
1. 7.4 × 10^–18
2. 6.7 × 10^–14
3. 1.4 × 10^–11
4. 7.2 × 10^–10
5. 7.0
Solution

Answer: C

Concept: K_a and K_b Relationship

Find K_b

Since we are at 25 °C, we know K_w = 1.0 × 10^–14.

\[\begin{align*} K_{\mathrm{a}} \times K_{\mathrm{b}} &= K_{\mathrm{w}} \longrightarrow \\[1.5ex] K_{\mathrm{b}} &= \dfrac{K_{\mathrm{w}}}{K_{\mathrm{a}}} \\[1.5ex] &= \dfrac{1\times 10^{-14}}{7.2\times 10^{-4}}\\[1.5ex] &= 1.389 \times 10^{-11} \end{align*}\]
Which acid is the strongest? Values are given at 25 °C.
1. HNO₂, pK_a = 3.39
2. HF, pK_a = 3.14
3. HCN, pK_a = 9.21
4. CH₃COOH, K_a = 1.8 × 10^–5
5. H₂CO₃, K_a = 4.3 × 10^–7
Solution

Answer: B

Concept: K_a and pK_a Relationship

Recall that the stronger the acid, the larger the K_a and the smaller the pK_a. Convert all values to either a K_a or pK_a and then determine the strongest acid.

The equations to convert are

\[\begin{align*} \mathrm{p}K_{\mathrm{a}} &= -\log K_{\mathrm{a}} \\[1.5ex] K_{\mathrm{a}} &= 10^{-\mathrm{p}K_{\mathrm{a}}} \end{align*}\]

List Everything as a K_a

HNO₂, K_a = 4.07 × 10^–4
HF, K_a = 7.24 × 10^–4
HCN, K_a = 6.16 × 10^–10
CH₃COOH, K_a = 1.8 × 10^–5
H₂CO₃, K_a = 4.3 × 10^–7

The largest K_a corresponds to HF. Therefore, HF is the strongest acid.

List Everything as pK_a

HNO₂, pK_a = 3.39
HF, pK_a = 3.14
HCN, pK_a = 9.21
CH₃COOH, pK_a = 4.74
H₂CO₃, pK_a = 6.37

The smallest pK_a corresponds to HF. Therefore, HF is the strongest acid.
What is [H₃O⁺] (in M) for a sample of pure water at 45 °C? K_w = 3.94 × 10^–14
1. 3.41 × 10^–8
2. 8.33 × 10^–4
3. 4.21 × 10⁶
4. 6.70
5. 1.98 × 10^–7
Solution

Answer: E

Concept: K_w

Remember that the pH of water changes with temperature because K_w changes! Recall the auto-ionization of water reaction

\[\mathrm{2H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH}^-(aq)\]

where the equilibrium expression is

\[K_{\mathrm{w}} = [\mathrm{H_3O^+}][\mathrm{OH^-}]\]

K_{} is only equal to 1.0 × 10^–14 at 25 °C. In this problem, water is at 45 °C and the K_w is given.

We can convert K_w to pK_w

\[\begin{align*} \mathrm{p}K_{\mathrm{w}} &= -\log K_{\mathrm{w}} \\[1.5ex] &= -\log 3.94 \times 10^{-14} \\[1.5ex] &= 13.408 \end{align*}\]

and then use

\[\mathrm{p}K_{\mathrm{w}} = \mathrm{pH} + \mathrm{pOH}\]

to get the pH. Since we’re considering pure water, we know that pH = pOH.

\[\begin{align*} \mathrm{p}K_{\mathrm{w}} &= \mathrm{pH} + \mathrm{pOH} \\[1.5ex] &= 2(\mathrm{pH}) \longrightarrow \\[1.5ex] \mathrm{pH} &= \dfrac{\mathrm{p}K_{\mathrm{w}}}{2} \\[1.5ex] &= \dfrac{13.408}{2} \\[1.5ex] &= 6.704 \end{align*}\]

Finally, convert pH into [H₃O⁺].

\[\begin{align*} [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 10^{-\mathrm{pH}} \\[1.5ex] &= 10^{-\mathrm{6.704}} \\[1.5ex] &= 1.977 \times 10^{-7} \end{align*}\]
The pOH of an aqueous solution is 6.22 (at 25 °C). What is the pH of the solution?
1. 6.44
2. 6.84
3. 7.00
4. 7.78
5. 8.02
Solution

Answer: D

Concept: pH and pOH

Recall that

\[\mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_{\mathrm{w}}\]

At 25 °C, pK_w = 14. Therefore,

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \longrightarrow \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 - 6.22 \\[1.5ex] &= 7.78 \end{align*}\]
A 0.15 M monoprotic acid is dissolved in pure water and the pH of solution is 2.43. What is the percent ionization of the acid?
1. 0.11%
2. 2.48%
3. 6.31%
4. 8.44%
5. 12.86%
Solution

Answer: B

Concept: Percent Ionization

Find Hydronium Ion Concentration at Equilibrium

Remember that a pH is directly related to a hydronium ion concentration at equilibrium.

\[\begin{align*} [\mathrm{H_3O^+}]_{\mathrm{eq}} &= 10^{-\mathrm{pH}} \\[1.5ex] &= 10^{-2.43} \\[1.5ex] &= 3.715\times 10^{-3}~M \end{align*}\]

Find Percent Ionization

\[\begin{align*} \%~\mathrm{ionization} &= \dfrac{[\mathrm{H_3O^+}]_{\mathrm {eq}}}{[\mathrm{HA}]_0} \times 100\% \\[1.5ex] &= \dfrac{3.715\times 10^{-3}~M}{0.15~M} \times 100\% \\[1.5ex] &= 2.477\% \end{align*}\]
Which of the following conjugates will react with water to affect the pH of solution?
1. CO₃^2–
2. Cl^–
3. NO₃^–
4. K⁺
5. Br^–
Solution

Answer: A

Concept: Salt Hydrolysis

We need to identify a substance in the list that is a conjugate to something that is “weak” since it will react with water to alter the pH of solution.

CO₃^2– is a conjugate to HCO₃^–, a weak acid.

Cl^– is a conjugate to HCl, a strong acid.

NO₃^– is a conjugate to HNO₃, a strong acid.

K⁺ is a Group 1A metal and does not react with water to change the pH of solution.

Br^– is a conjugate to HBr, a strong acid.
Which of the following conjugate bases does not react with water to produce OH^–?
1. CO₃^2–
2. F^–
3. CH₃COO^–
4. NO₂^–
5. I^–
Solution

Answer: E

Concept: Salt Hydrolysis

We need to identify a substance in the list that is a conjugate to something that is “strong” or a “Group 1A or 2A metal” since they will not react with water to alter the pH of solution.

CO₃^2– is a conjugate to HCO₃^–, a weak acid.

F^– is a conjugate to HF, a weak acid.

CH₃COOH^– is a weak acid.

NO₂^– is a conjugate to HNO₂, a weak acid.

I^– is a conjugate to HI, a strong acid.
Which salt, when dissolved in water, will lead to a basic solution?
1. KF
2. NaCl
3. NH₄Cl
4. LiNO₃
5. none of these
Solution

Answer: A

Concept: Salt Hydrolysis

We need to identify a substance in the list that is a weak base or contains a conjugate to a weak acid.

KF contains F^–, a conjgate base to a weak acid.

NaCl is a neutral salt.

NH₄Cl contains NH₄⁺, a weak acid.

LiNO₃ contains Li (a Group 1A metal – neutral) and NO₃^–, a conjugate to a strong acid.
Small, high-charge metal ions such as Al³⁺ and Fe³⁺ form what kind of solution when dissolved in water?
1. acidic
2. basic
3. neutral
Solution

Answer: A

Concept: Acidic Metal Hydrides
A 0.300 M HF aqueous solution is made (at 25 °C). What is the pH? K_a = 7.2 × 10^–4
1. 1.83
2. 2.48
3. 2.93
4. 3.15
5. 7.49
Solution

Answer: A

Concept: pH of a Weak Acid Solution

Math String: -log(sqrt(7.2e-4*.3))
What is the pOH of an aqueous solution containing 0.040 M HI (at 25 °C)?
1. 1.40
2. 3.24
3. 7.68
4. 8.43
5. 12.6
Solution

Answer: E

Concept: pH and pOH of a Strong Acid Solution

This is a strong acid solution. All the strong acid will become hydronium in water.

\[\mathrm{HI}(aq) + \mathrm{H_2O}(l) \longrightarrow \mathrm{H_3O^+}(aq) + \mathrm{I^-}(aq)\]

Therefore, the hydronium ion concentration is approximately equal to the concentration of the strong acid.

\[\mathrm{[HI]} \approx \mathrm{[H_3O^+]}\]

Find pH

Take the negative log of the strong acid molar concentration to get the pH.

\[\begin{align*} \mathrm{pH} &= -\log[\mathrm{H_3O^+}] \\[1.5ex] &= -\log(0.04~M) \\[1.5ex] &= 1.398 \end{align*}\]

Find the pOH

Subtract the pH from pK_w. At 25 °C, pK_w = 14.

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \longrightarrow \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 - 1.398 \\ &= 12.60 \end{align*}\]
What is the pH of a 0.05 M Ca(OH)₂ aqueous solution (at 25 °C)?
1. 6
2. 8
3. 10
4. 12
5. 13
Solution

Answer: E

Concept: pH of a Strong Base Solution

This is a strong base solution. All the salt will dissociate in water.

\[\mathrm{Ca(OH)_2}(s) \overset{\mathrm{H_2O}}\longrightarrow \mathrm{Ca^{2+}}(aq) + 2\mathrm{OH^-}(aq)\]

Find pOH

Recognize that the hydroxide ion concentration is twice that of the salt concentration due to stoichiometry.

\[\begin{align*} \mathrm{[OH^-]} &= 2[\mathrm{Ca(OH)_2}] \\[1.5ex] &= 2 (0.05~M) \\[1.5ex] &= 0.1~M \end{align*}\]

\[\begin{align*} \mathrm{pOH} &= -\log[\mathrm{OH^-}] \\[1.5ex] &= -\log(0.1~M) \\[1.5ex] &= 1 \end{align*}\]

Find the pH

Subtract the pOH from pK_w. At 25 °C, pK_w = 14.

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \longrightarrow \\[1.5ex] \mathrm{pH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pOH} \\[1.5ex] &= 14 - 1 \\ &= 13 \end{align*}\]
The equilibrium constant, K_b, corresponds to which of the following reactions?
1. NaCl(s) ⇌ Na⁺(aq) + Cl^–(aq)
2. F^–(aq) + H₂O(l) ⇌ HF(aq) + OH^–(aq)
3. NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)
4. HI(aq) ⇌ H⁺(aq) + I^–(aq)
5. none of these
Solution

Answer: B

Concept: Reaction Types and Equilibrium Constants

K_b is associated with a base-ionization reaction.
What is the molar concentration of a 1 L Ba(OH)₂ solution with a pH of 8.15 (at 25°C)? Be sure to account for the hydroxide ion contribution from water.
1. 6.56 × 10^–7
2. 1.31 × 10^–6
3. 1.41 × 10^–6
4. 8.34 × 10^–5
5. 4.28 × 10^–4
Solution

Answer: A

Concept: pH of a Strong Base Solution

Find pOH

Subtract the pH from pK_w. At 25 °C, pK_w = 14.

\[\begin{align*} \mathrm{pH} + \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} \longrightarrow \\[1.5ex] \mathrm{pOH} &= \mathrm{p}K_{\mathrm{w}} - \mathrm{pH} \\[1.5ex] &= 14 - 8.15 \\ &= 5.85 \end{align*}\]

Find [OH^–]

\[\begin{align*} [\mathrm{OH^-}] &= 10^{-\mathrm{pOH}} \\[1.5ex] &= 10^{-5.85} \\[1.5ex] &= 1.4125 \times 10^{-6}~M \end{align*}\]

Find [OH^–] from Ba(OH)₂

The hydroxide ion concentration from above accounts for all the hydroxide in solution. We must find the hydroxide ion concentration from the salt, Ba(OH)₂ by subtracting off the hydroxide ion concentration that comes from the water. We know that pure water has [OH^–] = 1 × 10^{–7. Therefore,}

\[\begin{align*} [\mathrm{OH^-}]_{\mathrm{Ba(OH)_2}} &= [\mathrm{OH^-}]_{\mathrm{tot}} - [\mathrm{OH^-}]_{\mathrm{H_2O}} \\[1.5ex] &= 1.4125 \times 10^{-6}~M - 1\times 10^{-7}~M \\[1.5ex] &= 1.3125 \times 10^{-6}~M \end{align*}\]

Find [Ba(OH)₂]

Now that we know how much hydroxide comes from the dissolved salt, use stoichiometry to determine the molar concentration of the salt. Here, for every one Ba(OH)₂ dissolved, two hydroxides are produced.

\[\mathrm{Ba(OH)_2}(s) \overset{\mathrm{H_2O}}\longrightarrow \mathrm{Ba^{2+}}(aq) + 2\mathrm{OH^-}(aq)\]

Therefore,

\[\begin{align*} [\mathrm{Ba(OH)_2}] &= \dfrac{1}{2}\left ( [\mathrm{OH^-}] \right ) \\[1.5ex] &= \dfrac{1}{2}\left ( 1.3125\times 10^{-6}~M \right ) \\[1.5ex] &= 6.56\times 10^{-7}~M \end{align*}\]
Which of the following aqueous mixtures will yield a buffer solution? (Assume all salts are soluble)
1. 1.05 M NH₃ + 1.35 M NH₄Cl
2. 0.80 M LiOH + 0.81 M LiNO₃
3. 1.43 M NaCl + 1.03 M NaOH
4. 0.5 M HBr + 0.5 M NaBr
5. none of these
Solution

Answer: A

Concept: Buffers

A buffer contains acid + conjugate base (or base + conjugate acid) in a 1:10 or 10:1 ratio. Answer A gives a solution containing a weak base (NH₃) and its conjugate acid (NH₄⁺) which comes from the ammonium chloride salt. Taking the ratio of base/acid gives

\[\dfrac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]} = \dfrac{1.35}{1.05} = 1.28\]

The ratio lies between 0.1 and 10. Therefore, answer A is a buffer.
Solution B is not a buffer since it does not contain a conjugate acid/base pair. Solution C is not a buffer because it contains a strong base and a neutral salt. Solution D is not a buffer because it contains a strong acid and a salt containing a conjugate to the strong acid.
What is the pH of a 1 L aqueous solution containing 0.02 mol HA and 0.015 mol A^– (at 25 °C)? K_a(HA) = 1.77 × 10^–4
1. 2.13
2. 2.59
3. 3.63
4. 3.98
5. 4.18
Solution

Answer: C

Concept: Buffers

A buffer contains acid + conjugate base (or base + conjugate acid) in a 1:10 or 10:1 ratio. Here, the ratio is 0.75 (or 1.333) giving us a buffer. Use Henderson-Hasselbalch to solve for the pH.

\[\begin{align*} \mathrm{pH} &= \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}} \\[1.5ex] &= -\log(1.77\times 10^{-4}) + \log \dfrac{0.015}{0.02} \\[1.5ex] &= 3.627 \end{align*}\]
A titration is carried out where a strong base is added to a weak acid solution. Choose the pH that best represents where the equivalence point would be. (Assume 25 °C)
1. 2.4
2. 5.4
3. 7.0
4. 8.5
5. 13.2
Solution

Answer: D

Concept: Weak Acid/Strong Base Titration

A weak acid/strong base titration should result in an equivalence point that is slightly basic (above a pH of 7 at 25 °C).
What is the pH of a 1 L buffer solution (at 25 °C) prepared from 0.15 mol of NH₃ and 0.20 mol NH₄I? K_a(NH₄⁺) = 5.8 × 10^–10
1. 4.34
2. 5.89
3. 6.22
4. 7.94
5. 9.11
Solution

Answer: E

Concept: Buffers

A buffer contains acid + conjugate base (or base + conjugate acid) in a 1:10 or 10:1 ratio. Use Henderson-Hasselbalch to solve for the pH.

\[\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\dfrac{[\mathrm{A^-}]}{\mathrm{[HA]}}\]
Which of the following K_b values (at 25 °C) represents the strongest base?
1. 6.5 × 10^–12
2. 1.7 × 10^–9
3. 1.6 × 10^–6
4. 2.1 × 10^–3
5. 6.4 × 10^–1
Solution

Answer: E

Concept: K_b and pK_b Relationship

Recall that the stronger the base, the larger the K_b.
What is the name for a reaction where an acid reacts with a base?
1. neutralization
2. combination
3. dissolution
4. dissociation
Solution

Answer: A

Concept: Reaction Name
Which equilibrium constant best corresponds with the following reaction?

\[\mathrm{2H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq)\]
1. K_a
2. K_b
3. K_w
4. K_sp
5. K_f
Solution

Answer: C

Concept: Reaction Types and Equilibrium Constants

K_w is associated with the auto-ionization of water reaction.
A titration is performed where 10.0 mL of a 0.70 M NaOH aqueous solution is added to 25.0 mL of a 0.150 M aqueous HCl solution. What is the final pH (at 25 °C)?
1. 1.01
2. 2.49
3. 12.97
4. 13.58
5. 14.34
Solution

Answer: C

Concept: Strong Acid/Strong Base Titration