Mastering Ionic Equilibrium
This Ionic Equilibrium Practice Quiz focuses on two critical areas of chemical equilibrium: buffer solutions and solubility product constants (Ksp). These concepts are essential for understanding how solutions resist pH changes and how sparingly soluble salts behave in aqueous environments.
Key Learning Objectives:
- Apply the Henderson-Hasselbalch equation to calculate the pH of acidic and basic buffers.
- Determine the molar solubility of salts from their ##K_{sp}## values and vice versa.
- Predict the formation of precipitates using the reaction quotient (##Q_{sp}##).
- Evaluate the impact of the common ion effect on solubility and equilibrium.
This moderate-level quiz is designed to challenge your understanding of multi-step reasoning and quantitative chemical analysis.
Q1. A buffer solution is prepared by mixing 0.20 M ##CH_3COOH## and 0.10 M ##CH_3COONa##. If the ##pK_a## of acetic acid is 4.74, what is the pH of the resulting solution?
Using the Henderson-Hasselbalch equation: ##pH = pK_a + \log([Salt]/[Acid])##. Substituting the values: ##pH = 4.74 + \log(0.10/0.20) = 4.74 + \log(0.5) = 4.74 - 0.30 = 4.44##.
Q2. The solubility product constant (##K_{sp}##) for silver chloride (##AgCl##) is ##1.8 \times 10^{-10}## at 25°C. What is the molar solubility of ##AgCl## in pure water?
For a binary salt like ##AgCl##, ##K_{sp} = s^2##, where ##s## is the molar solubility. Therefore, ##s = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} M##.
Q3. A buffer solution contains 0.50 M ammonia (##NH_3##) and 0.50 M ammonium chloride (##NH_4Cl##). If the ##pK_b## of ammonia is 4.75, what is the pH of this buffer?
First calculate pOH: ##pOH = pK_b + \log([Salt]/[Base]) = 4.75 + \log(0.50/0.50) = 4.75 + 0 = 4.75##. Then, ##pH = 14 - pOH = 14 - 4.75 = 9.25##.
Q4. Which of the following changes will occur if a small amount of ##HCl## is added to a buffer solution consisting of ##HF## and ##NaF##?
When ##H^+## from ##HCl## is added, it reacts with the conjugate base ##F^-## to form more weak acid ##HF## (##H^+ + F^- \rightarrow HF##), thus increasing the concentration of ##HF##.
Q5. The molar solubility of ##CaF_2## in water is ##s##. Which expression correctly relates ##s## to the solubility product constant ##K_{sp}##?
The dissociation is ##CaF_2 \rightleftharpoons Ca^{2+} + 2F^-##. If solubility is ##s##, then ##[Ca^{2+}] = s## and ##[F^-] = 2s##. ##K_{sp} = [Ca^{2+}][F^-]^2 = (s)(2s)^2 = 4s^3##.
Q6. Calculate the pH of a buffer solution that is 0.15 M in ##NH_3## and 0.35 M in ##NH_4Cl##. (##K_b## for ##NH_3 = 1.8 \times 10^{-5}##).
##pK_b = -\log(1.8 \times 10^{-5}) = 4.74##. ##pOH = 4.74 + \log(0.35/0.15) = 4.74 + 0.37 = 5.11##. ##pH = 14 - 5.11 = 8.89##.
Q7. What is the molar solubility of ##Mg(OH)_2## (##K_{sp} = 1.8 \times 10^{-11}##) in a solution where the pH is maintained at 12.00?
If pH = 12, pOH = 2, so ##[OH^-] = 10^{-2} M##. ##K_{sp} = [Mg^{2+}][OH^-]^2 \Rightarrow 1.8 \times 10^{-11} = [Mg^{2+}](0.01)^2##. Thus, ##[Mg^{2+}] = (1.8 \times 10^{-11}) / 10^{-4} = 1.8 \times 10^{-7} M##.
Q8. Under what condition will a precipitate of ##Ag_2SO_4## form when mixing solutions of ##AgNO_3## and ##Na_2SO_4##?
Precipitation occurs only when the ion product (##Q_{sp}##) exceeds the solubility product constant (##K_{sp}##) of the salt.
Q9. A buffer is most effective at resisting pH changes when:
A buffer has its maximum buffering capacity when the concentrations of the weak acid and its conjugate base are equal, which occurs when ##pH = pK_a##.
Q10. Calculate the ##K_{sp}## for ##PbI_2## if its molar solubility in water is ##1.5 \times 10^{-3} M##.
For ##PbI_2##, ##K_{sp} = 4s^3##. Substituting ##s = 1.5 \times 10^{-3}##: ##K_{sp} = 4 \times (1.5 \times 10^{-3})^3 = 4 \times (3.375 \times 10^{-9}) = 1.35 \times 10^{-8}##.
Q11. Which of the following salt solutions would be the least soluble in 0.1 M ##HCl## compared to pure water?
The solubility of salts containing basic anions (like carbonate, hydroxide, or phosphate) increases in acidic solution (##HCl##) because the acid reacts with the anion. ##AgCl## solubility actually decreases in ##HCl## due to the common ion effect of ##Cl^-##.
Q12. To prepare a buffer with a pH of 9.0 using a weak base (##K_b = 1.0 \times 10^{-5}##) and its conjugate acid, what should be the ratio of ##[Salt]/[Base]##?
pH = 9.0 means pOH = 5.0. Given ##pK_b = -\log(10^{-5}) = 5.0##. Using ##pOH = pK_b + \log([Salt]/[Base])##, we get ##5.0 = 5.0 + \log([Salt]/[Base])##, so ##\log([Salt]/[Base]) = 0##, meaning the ratio is 1:1. Wait, let me re-check. If pH=9, pOH=5, pKb=5. Ratio is 1. If pH was 8, pOH=6, then ratio 10:1. Correction: If pH=9, pOH=5, pKb=5, ratio is 1. Let's adjust question for pH 8. If pH=8, pOH=6, ratio is 10:1.
Q13. What happens to the molar solubility of ##BaSO_4## when ##Na_2SO_4## is added to the solution?
This is the common ion effect. The addition of ##SO_4^{2-}## ions from ##Na_2SO_4## shifts the equilibrium ##BaSO_4(s) \rightleftharpoons Ba^{2+} + SO_4^{2-}## to the left, decreasing the solubility of ##BaSO_4##.
Q14. A solution has a concentration of ##0.010 M Ba^{2+}##. What concentration of ##SO_4^{2-}## is required to just begin the precipitation of ##BaSO_4##? (##K_{sp} = 1.1 \times 10^{-10}##).
Precipitation begins when ##Q_{sp} = K_{sp}##. ##[Ba^{2+}][SO_4^{2-}] = 1.1 \times 10^{-10} \Rightarrow (0.010)[SO_4^{2-}] = 1.1 \times 10^{-10} \Rightarrow [SO_4^{2-}] = 1.1 \times 10^{-8} M##.
Q15. If the molar solubility of a salt ##A_2B## is ##2.0 \times 10^{-4} M##, what is its ##K_{sp}##?
For ##A_2B##, ##K_{sp} = 4s^3##. Substituting ##s = 2.0 \times 10^{-4}##: ##K_{sp} = 4 \times (2.0 \times 10^{-4})^3 = 4 \times (8.0 \times 10^{-12}) = 3.2 \times 10^{-11}##.
Q16. Which pair of compounds can be used to prepare a buffer solution?
A buffer solution must consist of a weak acid and its conjugate base, or a weak base and its conjugate acid. ##NH_3## (weak base) and ##NH_4Cl## (conjugate acid) fit this definition.
Q17. The pH of a buffer solution containing 0.1 M weak acid (##HA##) and 0.1 M conjugate base (##A^-##) is 5.0. What is the ##K_a## of the acid?
In a buffer where ##[Acid] = [Base]##, ##pH = pK_a##. If ##pH = 5.0##, then ##pK_a = 5.0##. Since ##K_a = 10^{-pK_a}##, ##K_a = 10^{-5}##.
Q18. What is the molar solubility of ##AgBr## (##K_{sp} = 5.0 \times 10^{-13}##) in a 0.10 M ##NaBr## solution?
In 0.10 M ##NaBr##, ##[Br^-] = 0.10 M##. ##K_{sp} = [Ag^+][Br^-] \Rightarrow 5.0 \times 10^{-13} = [Ag^+](0.10)##. Thus, ##[Ag^+] = 5.0 \times 10^{-12} M##, which is the molar solubility.
Q19. Consider the equilibrium: ##AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)##. Adding which substance will increase the solubility of ##AgCl##?
Adding ##NH_3## increases the solubility of ##AgCl## because ##NH_3## reacts with ##Ag^+## to form a stable complex ion ##[Ag(NH_3)_2]^+##, effectively removing ##Ag^+## from the equilibrium and shifting it to the right.
Q20. A student mixes 100 mL of 0.01 M ##Pb(NO_3)_2## with 100 mL of 0.02 M ##KI##. Given ##K_{sp}## for ##PbI_2## is ##1.4 \times 10^{-8}##, will a precipitate form?
After mixing, concentrations are halved: ##[Pb^{2+}] = 0.005 M##, ##[I^-] = 0.01 M##. ##Q_{sp} = [Pb^{2+}][I^-]^2 = (0.005)(0.01)^2 = 5 \times 10^{-7}##. Since ##5 \times 10^{-7} > 1.4 \times 10^{-8}##, a precipitate forms.
Q21. What is the pH of a solution prepared by mixing 50 mL of 0.1 M ##NaOH## with 100 mL of 0.1 M ##CH_3COOH## (##pK_a = 4.74##)?
##NaOH## reacts with ##CH_3COOH## to produce ##CH_3COONa##. Moles ##NaOH = 0.005##, Moles ##Acid = 0.01##. After reaction: Moles ##Salt = 0.005##, Moles ##Acid remaining = 0.005##. Since ##[Salt] = [Acid]##, ##pH = pK_a = 4.74##.
Q22. The solubility of ##MgF_2## is ##1.2 \times 10^{-3} M##. Calculate its solubility product constant ##K_{sp}##.
For ##MgF_2##, ##K_{sp} = 4s^3##. ##K_{sp} = 4 \times (1.2 \times 10^{-3})^3 = 4 \times (1.728 \times 10^{-9}) = 6.912 \times 10^{-9}##.
Q23. Which of the following buffers would be best for maintaining a pH of 3.5?
A buffer is most effective when the desired pH is within ##\pm 1## unit of the ##pK_a##. Formic acid with ##pK_a = 3.74## is the closest to 3.5.
Q24. What is the effect of diluting a buffer solution with an equal volume of pure water?
Dilution changes the concentrations of both the weak acid and conjugate base equally, so their ratio remains the same. According to Henderson-Hasselbalch, the pH remains largely unchanged, though the buffer capacity decreases.
Q25. If the ##K_{sp}## of ##Ca(OH)_2## is ##5.5 \times 10^{-6}##, what is the pH of a saturated solution of calcium hydroxide?
##K_{sp} = 4s^3 = 5.5 \times 10^{-6} \Rightarrow s^3 = 1.375 \times 10^{-6} \Rightarrow s \approx 0.0111 M##. Since ##[OH^-] = 2s = 0.0222 M##, ##pOH = -\log(0.0222) \approx 1.65##. ##pH = 14 - 1.65 = 12.35##.
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