7.13 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.
What is the Ka 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) + H2O(l) ⇌ 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 Ka.
\[\begin{align*} K_{\mathrm{a}} &= \dfrac{[\mathrm{H_3O}^+][\mathrm{A^-}]}{[\mathrm{HA}]}\\[1.5ex] &= \dfrac{(1.41 \times 10^{-4}~M)(1.41 \times 10^{-4}~M)}{0.049859~M}\\[1.5ex] &= 3.987\times 10^{-7} \end{align*}\]
What is the pH of water at 60 °C? Kw = 9.311 × 10–14
- 6.5
- 6.9
- 7.0
- 7.4
- 8.2
Solution
Answer: A
Concept: Kw
Remember that the pH of water changes with temperature because Kw 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
\[K_{\mathrm{w}} = [\mathrm{H_3O^+}][\mathrm{OH^-}]\]
Kw is only equal to 1.0 × 10–14 at 25 °C. In this problem, water is at 60 °C and the Kw is given.
We can convert Kw to pKw
\[\begin{align*} \mathrm{p}K_{\mathrm{w}} &= -\log K_{\mathrm{w}} \\[1.5ex] &= -\log 9.311\times 10^{-14} \\[1.5ex] &= 13.031 \end{align*}\]
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)?
- 1.40
- 3.24
- 7.68
- 8.43
- 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 Ka 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 Kw = 1.0 × 10–14 and, therefore, pKw = 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.
\[\mathrm{NO_3}^-(aq) + \mathrm{H_3O^+}(aq) \rightleftharpoons \mathrm{H_2O}(l) + \mathrm{HNO_3}(aq)\]
- H2O
- HNO3
- H3O+
- NO3–
- 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?
- Arrhenius base
- Lewis base
- Brønsted-Lowry acid
- Lewis acid
Solution
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)?
- 0.3
- 0.6
- 1.2
- 2.4
- 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 Ka 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 Ka for hydrofluoric acid is 7.2 × 10–4 at 25 °C. What is the corresponding Kb for its conjugate base?
- 7.4 × 10–18
- 6.7 × 10–14
- 1.4 × 10–11
- 7.2 × 10–10
- 7.0
Solution
Answer: C
Concept: Ka and Kb Relationship
Find Kb
Since we are at 25 °C, we know Kw = 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.
- HNO2, pKa = 3.39
- HF, pKa = 3.14
- HCN, pKa = 9.21
- CH3COOH, Ka = 1.8 × 10–5
- H2CO3, Ka = 4.3 × 10–7
Solution
Answer: B
Concept: Ka and pKa Relationship
Recall that the stronger the acid, the larger the Ka and the smaller the pKa. Convert all values to either a Ka or pKa 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 Ka
HNO2, Ka = 4.07 × 10–4
HF, Ka = 7.24 × 10–4
HCN, Ka = 6.16 × 10–10
CH3COOH, Ka = 1.8 × 10–5
H2CO3, Ka = 4.3 × 10–7The largest Ka corresponds to HF. Therefore, HF is the strongest acid.
List Everything as pKa
HNO2, pKa = 3.39
HF, pKa = 3.14
HCN, pKa = 9.21
CH3COOH, pKa = 4.74
H2CO3, pKa = 6.37The smallest pKa corresponds to HF. Therefore, HF is the strongest acid.
What is [H3O+] (in M) for a sample of pure water at 45 °C? Kw = 3.94 × 10–14
- 3.41 × 10–8
- 8.33 × 10–4
- 4.21 × 106
- 6.70
- 1.98 × 10–7
Solution
Answer: E
Concept: Kw
Remember that the pH of water changes with temperature because Kw 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 Kw is given.
We can convert Kw to pKw
\[\begin{align*} \mathrm{p}K_{\mathrm{w}} &= -\log K_{\mathrm{w}} \\[1.5ex] &= -\log 3.94 \times 10^{-14} \\[1.5ex] &= 13.408 \end{align*}\]
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?
- 6.44
- 6.84
- 7.00
- 7.78
- 8.02
Solution
Answer: D
Concept: pH and pOH
Recall that
\[\mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_{\mathrm{w}}\]
At 25 °C, pKw = 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?
- 0.11%
- 2.48%
- 6.31%
- 8.44%
- 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?
- CO32–
- Cl–
- NO3–
- K+
- 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.
CO32– 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–?
- CO32–
- F–
- CH3COO–
- NO2–
- 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.
CO32– 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?
- KF
- NaCl
- NH4Cl
- LiNO3
- 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?
- acidic
- basic
- neutral
Solution
Answer: A
Concept: Acidic Metal Hydrides
A 0.300 M HF aqueous solution is made (at 25 °C). What is the pH? Ka = 7.2 × 10–4
- 1.83
- 2.48
- 2.93
- 3.15
- 7.49
Solution
What is the pOH of an aqueous solution containing 0.040 M HI (at 25 °C)?
- 1.40
- 3.24
- 7.68
- 8.43
- 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 pKw. At 25 °C, pKw = 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)?
- 6
- 8
- 10
- 12
- 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 pKw. At 25 °C, pKw = 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, Kb, corresponds to which of the following reactions?
- NaCl(s) ⇌ Na+(aq) + Cl–(aq)
- F–(aq) + H2O(l) ⇌ HF(aq) + OH–(aq)
- NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
- HI(aq) ⇌ H+(aq) + I–(aq)
- none of these
Solution
Answer: B
Concept: Reaction Types and Equilibrium Constants
Kb 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.
- 6.56 × 10–7
- 1.31 × 10–6
- 1.41 × 10–6
- 8.34 × 10–5
- 4.28 × 10–4
Solution
Answer: A
Concept: pH of a Strong Base Solution
Find pOH
Subtract the pH from pKw. At 25 °C, pKw = 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)
- 1.05 M NH3 + 1.35 M NH4Cl
- 0.80 M LiOH + 0.81 M LiNO3
- 1.43 M NaCl + 1.03 M NaOH
- 0.5 M HBr + 0.5 M NaBr
- 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.- 1.05 M NH3 + 1.35 M NH4Cl
What is the pH of a 1 L aqueous solution containing 0.02 mol HA and 0.015 mol A– (at 25 °C)? Ka(HA) = 1.77 × 10–4
- 2.13
- 2.59
- 3.63
- 3.98
- 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)
- 2.4
- 5.4
- 7.0
- 8.5
- 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? Ka(NH4+) = 5.8 × 10–10
- 4.34
- 5.89
- 6.22
- 7.94
- 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 Kb values (at 25 °C) represents the strongest base?
- 6.5 × 10–12
- 1.7 × 10–9
- 1.6 × 10–6
- 2.1 × 10–3
- 6.4 × 10–1
Solution
What is the name for a reaction where an acid reacts with a base?
- neutralization
- combination
- dissolution
- dissociation
Solution
Answer: A
Concept: Reaction Name
Which equilibrium constant best corresponds with the following reaction?
\[\mathrm{2H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq)\]
- Ka
- Kb
- Kw
- Ksp
- Kf
Solution
Answer: C
Concept: Reaction Types and Equilibrium Constants
Kw 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)?
- 1.01
- 2.49
- 12.97
- 13.58
- 14.34
Solution
Answer: C
Concept: Strong Acid/Strong Base Titration