Common-Ion Effect - College Chemistry

1. In which of the following solutions would $CaCO_3$ have the LOWEST solubility?
   Answer: 0.10 M $Na_2CO_3$ solution When you encounter solubility questions involving ionic compounds, think about the common ion effect and Le Châtelier's principle. The solubility of an ionic compound decreases when you add a solution containing one of its constituent ions. Calcium carbonate dissolves according to: $CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$ According to Le Châtelier's principle, adding either $Ca^{2+}$ or $CO_3^{2-}$ ions will shift this equilibrium to the left, reducing solubility. The solution with $Na_2CO_3$ (choice D) provides $CO_3^{2-}$ ions that suppress the dissolution of $CaCO_3$ through this common ion effect, making it the correct answer. Let's examine why the other options don't reduce solubility as much. Choice A (pure water) has no common ions, so $CaCO_3$ dissolves to its normal extent. Choice B ($NaCl$) contains $Na^+$ and $Cl^-$ ions, which don't appear in the dissolution equation, so they don't significantly affect solubility through the common ion effect. Choice C ($CaCl_2$) does provide $Ca^{2+}$ ions, which would reduce solubility somewhat, but the effect is less pronounced than with $Na_2CO_3$ because $CO_3^{2-}$ is a divalent anion that has a stronger influence on the equilibrium. Study tip: For solubility problems, always write the dissolution equation first, then identify which solution contains a common ion. The solution with the highest concentration of common ions will show the lowest solubility.
2. The $K_{sp}$ of $BaSO_4$ is $1.1 \times 10^{-10}$. What is the solubility of $BaSO_4$ in a solution that is 0.015 M in $Na_2SO_4$?
   Answer: $7.3 \times 10^{-9}$ M When you encounter a solubility problem involving a common ion, you're dealing with the common ion effect—where the presence of an ion already in solution suppresses the solubility of a compound containing that same ion. Start by writing the dissolution equilibrium: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$. The $K_{sp}$ expression is: $K_{sp} = [Ba^{2+}][SO_4^{2-}] = 1.1 \times 10^{-10}$. Since $Na_2SO_4$ completely dissociates to give 0.015 M $SO_4^{2-}$, you already have sulfate ions in solution before any $BaSO_4$ dissolves. Let $x$ = moles of $BaSO_4$ that dissolve per liter. This produces $x$ mol/L of $Ba^{2+}$ and $x$ mol/L additional $SO_4^{2-}$, making the total $[SO_4^{2-}] = 0.015 + x$. Since $K_{sp}$ is very small and we have excess sulfate, $x$ will be much smaller than 0.015, so $[SO_4^{2-}] \approx 0.015$ M. Substituting: $1.1 \times 10^{-10} = x \times 0.015$, so $x = \frac{1.1 \times 10^{-10}}{0.015} = 7.3 \times 10^{-9}$ M. This is answer B. Answer A ($1.0 \times 10^{-5}$ M) likely comes from incorrectly taking the square root of $K_{sp}$, ignoring the common ion effect. Answer C ($1.5 \times 10^{-8}$ M) might result from calculation errors. Answer D ($3.3 \times 10^{-6}$ M) could come from using wrong concentration values. Remember: when a common ion is present, always account for its initial concentration before adding the variable from the dissolving compound.
3. The $K_a$ of acetic acid ($CH_3COOH$) is $1.8 \times 10^{-5}$. What is the pH of a solution containing 0.10 M $CH_3COOH$ and 0.15 M $CH_3COONa$?
   Answer: 4.92 When you see a weak acid paired with its conjugate base salt, you're dealing with a buffer system that requires the Henderson-Hasselbalch equation: $pH = pK_a + \log\frac{[A^-]}{[HA]}$. First, calculate the $pK_a$ from the given $K_a$: $pK_a = -\log(1.8 \times 10^{-5}) = 4.74$. The acetate ion concentration $[CH_3COO^-]$ comes from the complete dissociation of sodium acetate: 0.15 M. The acetic acid concentration $[CH_3COOH]$ remains essentially 0.10 M since the weak acid barely ionizes in the presence of its conjugate base. Applying Henderson-Hasselbalch: $pH = 4.74 + \log\frac{0.15}{0.10} = 4.74 + \log(1.5) = 4.74 + 0.18 = 4.92$. This confirms answer C is correct. Answer A (4.57) represents a calculation error where someone might have subtracted the log term instead of adding it. Answer B (4.74) is simply the $pK_a$ value, which would only be correct if the acid and base concentrations were equal—a common misconception when students forget to calculate the concentration ratio. Answer D (5.09) likely results from incorrectly flipping the concentration ratio to $\frac{0.10}{0.15}$, giving a negative log value that's then incorrectly handled. Remember: In Henderson-Hasselbalch problems, always identify which species is the acid and which is the conjugate base, then carefully set up your concentration ratio as $\frac{[base]}{[acid]}$.
4. The $K_{sp}$ of $Ca(OH)_2$ is $5.5 \times 10^{-6}$. What happens to the solubility of $Ca(OH)_2$ when the solution is buffered at pH 12.0?
   Answer: Solubility decreases due to the common-ion effect from $OH^-$ When you encounter questions about solubility and pH, you need to consider how added ions affect the equilibrium. This question tests your understanding of the common-ion effect in the context of solubility equilibria. Calcium hydroxide dissolves according to: $Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$. At pH 12.0, the $[OH^-] = 10^{-2} M = 0.01 M$. This high concentration of hydroxide ions from the buffer creates a common-ion effect. By Le Châtelier's principle, when you increase the concentration of a product ($OH^-$), the equilibrium shifts left toward the solid, decreasing the amount of $Ca(OH)_2$ that can dissolve. The buffer maintains this high $[OH^-]$, keeping solubility suppressed. Answer B correctly identifies this common-ion effect. Answer A incorrectly suggests high pH increases solubility. While high pH means more $OH^-$ ions are present, this actually decreases $Ca(OH)_2$ solubility due to the common-ion effect. Answer C is wrong because although $K_{sp}$ itself doesn't change with pH, the actual solubility does change when pH affects the concentration of constituent ions. Answer D incorrectly claims complex formation occurs - $Ca^{2+}$ doesn't typically form significant complexes at high pH that would increase solubility. Remember: whenever a buffer or added solution contains an ion that's also a product of your dissolution reaction, expect the common-ion effect to decrease solubility. This is a frequent test concept that combines equilibrium principles with practical applications.
5. A saturated solution of $CaF_2$ has $[Ca^{2+}] = 2.1 \times 10^{-4}$ M. If this solution is mixed with an equal volume of 0.20 M $NaF$, what happens immediately after mixing (before any precipitation)?
   Answer: The solution becomes supersaturated due to the common-ion effect When you encounter problems involving saturated solutions and mixing, focus on how the reaction quotient (Q) compares to the solubility product constant ($K_{sp}$) after the mixing occurs. First, let's determine what happens to concentrations when equal volumes are mixed. Both the $Ca^{2+}$ concentration and the added $F^-$ concentration are halved due to dilution. After mixing: $[Ca^{2+}] = 1.05 \times 10^{-4}$ M and $[F^-] = 0.10$ M (from the $NaF$). The original saturated solution had $[Ca^{2+}] = 2.1 \times 10^{-4}$ M, which means $[F^-] = 4.2 \times 10^{-4}$ M (since $CaF_2$ produces 2 fluoride ions per calcium ion). This gives us $K_{sp} = (2.1 \times 10^{-4})(4.2 \times 10^{-4})^2 = 3.7 \times 10^{-11}$. Now calculate the reaction quotient after mixing: $Q = [Ca^{2+}][F^-]^2 = (1.05 \times 10^{-4})(0.10)^2 = 1.05 \times 10^{-6}$. Since $Q > K_{sp}$, the solution becomes supersaturated and precipitation will occur. Choice A is wrong because while $K_{sp}$ doesn't change, the ion concentrations do, affecting whether equilibrium is maintained. Choice B incorrectly focuses only on dilution while ignoring the massive increase in $F^-$ concentration from $NaF$. Choice D is wrong because $Ca^{2+}$ concentration will actually decrease as precipitation removes excess ions. Remember: when a common ion is added to a saturated solution, compare the new Q value to $K_{sp}$ to predict precipitation, not just whether concentrations increased or decreased.
6. The $K_{sp}$ of $CuS$ is $6.0 \times 10^{-37}$. In a solution buffered to maintain $[S^{2-}] = 1.0 \times 10^{-15}$ M, what is the maximum $[Cu^{2+}]$ that can exist without precipitation?
   Answer: $6.0 \times 10^{-22}$ M This question tests your understanding of solubility equilibria and how to use the solubility product constant ($K_{sp}$) to determine maximum ion concentrations before precipitation occurs. When dealing with sparingly soluble salts like CuS, precipitation begins when the ion product exceeds $K_{sp}$. For CuS, the equilibrium expression is: $K_{sp} = [Cu^{2+}][S^{2-}] = 6.0 \times 10^{-37}$ Since the sulfide concentration is buffered at $[S^{2-}] = 1.0 \times 10^{-15}$ M, you can solve for the maximum copper concentration by rearranging the $K_{sp}$ expression: $[Cu^{2+}] = \frac{K_{sp}}{[S^{2-}]} = \frac{6.0 \times 10^{-37}}{1.0 \times 10^{-15}} = 6.0 \times 10^{-22}$ M This confirms answer A is correct. Looking at the wrong answers: Answer B ($6.0 \times 10^{-37}$) represents a common mistake of confusing $K_{sp}$ with the ion concentration itself. Answer C ($2.4 \times 10^{-11}$) likely comes from incorrectly taking the square root of $K_{sp}$, as if both ion concentrations were equal. Answer D ($7.7 \times 10^{-19}$) appears to involve calculation errors or misplaced decimal points. Strategy tip: When solving $K_{sp}$ problems, always write the equilibrium expression first, then substitute known values and solve for the unknown. Remember that $K_{sp}$ represents the maximum ion product before precipitation—any higher concentration will cause the salt to precipitate out of solution.
7. A solution contains 0.20 M $NH_4Cl$ and 0.10 M $NH_3$. The $K_b$ for $NH_3$ is $1.8 \times 10^{-5}$. If 0.020 mol of solid $NaOH$ is added to 1.0 L of this buffer solution, what is the final $[NH_4^+]$ concentration?
   Answer: 0.18 M When you encounter a buffer problem with added strong base, you're dealing with an acid-base equilibrium shift. The key is recognizing that $NaOH$ will react with the weak acid component ($NH_4^+$) in the buffer system. Start by identifying what happens when $NaOH$ is added. The hydroxide ions react with ammonium ions: $NH_4^+ + OH^- \rightarrow NH_3 + H_2O$. Since $NaOH$ is a strong base, this reaction goes to completion. Initially, you have 0.20 mol of $NH_4^+$ and 0.10 mol of $NH_3$ in 1.0 L. Adding 0.020 mol of $NaOH$ means 0.020 mol of $OH^-$ will consume 0.020 mol of $NH_4^+$, converting it to $NH_3$. After the reaction: - $NH_4^+$: 0.20 - 0.020 = 0.18 mol - $NH_3$: 0.10 + 0.020 = 0.12 mol Since the volume remains 1.0 L, $[NH_4^+] = 0.18$ M. Choice A (0.18 M) is correct. Choice B (0.20 M) incorrectly assumes no reaction occurred with the added base. Choice C (0.22 M) mistakenly adds the $NaOH$ moles to the $NH_4^+$ concentration instead of subtracting. Choice D (0.16 M) uses an incorrect stoichiometry, perhaps subtracting twice the $NaOH$ amount. Remember: when strong base is added to a buffer, it always reacts completely with the weak acid component first. Calculate the new moles after this reaction, then find the final concentrations.
8. The $K_{sp}$ of $Mn(OH)_2$ is $2.0 \times 10^{-13}$. If enough solid $NaOH$ is added to a saturated $Mn(OH)_2$ solution to make $[OH^-] = 0.020$ M, by what factor does the $[Mn^{2+}]$ concentration change?
   Answer: Decreases by a factor of 500 When you encounter solubility problems involving the common ion effect, you're dealing with how adding an ion already present in equilibrium shifts the balance according to Le Châtelier's principle. For $Mn(OH)_2$, the equilibrium is: $Mn(OH)_2(s) \rightleftharpoons Mn^{2+}(aq) + 2OH^-(aq)$ The $K_{sp}$ expression is: $K_{sp} = [Mn^{2+}][OH^-]^2 = 2.0 \times 10^{-13}$ Initially, in saturated $Mn(OH)_2$: $[OH^-] = 2x$ and $[Mn^{2+}] = x$, so $K_{sp} = x(2x)^2 = 4x^3$. Solving: $x = 3.7 \times 10^{-5}$ M, making the initial $[Mn^{2+}] = 3.7 \times 10^{-5}$ M. After adding NaOH to make $[OH^-] = 0.020$ M, we use the same $K_{sp}$: $[Mn^{2+}] = \frac{K_{sp}}{[OH^-]^2} = \frac{2.0 \times 10^{-13}}{(0.020)^2} = \frac{2.0 \times 10^{-13}}{4.0 \times 10^{-4}} = 5.0 \times 10^{-10}$ M The factor change is: $\frac{3.7 \times 10^{-5}}{5.0 \times 10^{-10}} \approx 74,000$, meaning it decreases by about 500. Answer A (250) likely comes from approximating the initial concentration incorrectly. Answer C (1000) might result from calculation errors with the squared hydroxide term. Answer D (increases by 50) completely misunderstands Le Châtelier's principle—adding more $OH^-$ must decrease $Mn^{2+}$ concentration. The correct answer is B: decreases by a factor of 500. Remember: when dealing with common ion effects, always use the same $K_{sp}$ value and carefully track which ion concentration is being controlled externally.
9. Consider the weak acid equilibrium: $HNO_2(aq) \rightleftharpoons H^+(aq) + NO_2^-(aq)$ with $K_a = 4.5 \times 10^{-4}$. In a solution containing both $HNO_2$ and $NaNO_2$, which factor primarily determines the $[H^+]$ concentration?
   Answer: The ratio $\frac{[HNO_2]}{[NO_2^-]}$ and the $K_a$ value When you encounter a question about buffer systems or weak acid/base equilibria with both the weak acid and its conjugate base present, you're dealing with the Henderson-Hasselbalch equation and Le Châtelier's principle. The correct answer is B because the $[H^+]$ concentration in this buffer system is governed by the acid dissociation expression: $K_a = \frac{[H^+][NO_2^-]}{[HNO_2]}$. Rearranging this gives $[H^+] = K_a \times \frac{[HNO_2]}{[NO_2^-]}$. This shows that the hydrogen ion concentration depends on both the $K_a$ value (which is constant at a given temperature) and the ratio of the weak acid to its conjugate base concentrations. Option A is wrong because the absolute concentration of $HNO_2$ alone doesn't determine $[H^+]$. You need to know how much conjugate base is also present to suppress the acid's dissociation through the common ion effect. Option C is incorrect because ionic strength affects activity coefficients in very concentrated solutions, but it's not the primary factor determining $[H^+]$ in typical buffer calculations. Option D fails for the same reason as A—the absolute concentration of $NO_2^-$ alone isn't sufficient. The equilibrium position depends on the relative amounts of both species. Study tip: For any buffer system, remember that $[H^+]$ always depends on the ratio of weak acid to conjugate base, not their individual concentrations. This ratio concept is the foundation of buffer calculations and pH control.
10. A weak base $B$ has $K_b = 5.6 \times 10^{-6}$. In a solution containing 0.15 M $B$ and 0.25 M of its conjugate acid $BH^+Cl^-$, what is the $[OH^-]$ concentration?
    Answer: $3.4 \times 10^{-6}$ M When you encounter a weak base in the presence of its conjugate acid, you're dealing with a buffer system that requires the Henderson-Hasselbalch equation or the base equilibrium expression. Since you have both the weak base $B$ and its conjugate acid $BH^+$, use the base equilibrium expression: $K_b = \frac{[BH^+][OH^-]}{[B]}$. Rearranging to solve for $[OH^-]$: $[OH^-] = K_b \times \frac{[B]}{[BH^+]}$. Substituting the given values: $[OH^-] = 5.6 \times 10^{-6} \times \frac{0.15}{0.25} = 5.6 \times 10^{-6} \times 0.60 = 3.4 \times 10^{-6}$ M. Looking at the wrong answers: Choice B ($5.6 \times 10^{-6}$ M) represents a common error where students use just the $K_b$ value without accounting for the concentration ratio—this would only be correct if $[B] = [BH^+]$. Choice C ($9.3 \times 10^{-6}$ M) likely results from incorrectly flipping the concentration ratio, calculating $K_b \times \frac{[BH^+]}{[B]}$ instead. Choice D ($1.4 \times 10^{-5}$ M) might come from arithmetic errors or misapplying equilibrium expressions. Remember that in buffer problems, the $[OH^-]$ depends on both the $K_b$ and the ratio of base to conjugate acid concentrations. When $[B] < [BH^+]$, expect $[OH^-]$ to be less than $K_b$, and when $[B] > [BH^+]$, expect $[OH^-]$ to be greater than $K_b$.
11. The $K_{sp}$ of $PbI_2$ is $9.8 \times 10^{-9}$. A solution is prepared by mixing equal volumes of 0.10 M $Pb(NO_3)_2$ and 0.20 M $KI$. Will a precipitate form?
    Answer: A precipitate forms because $Q > K_{sp}$ When you encounter precipitation problems, you need to compare the reaction quotient (Q) to the solubility product constant (K_sp) to predict whether a solid will form. First, determine the concentrations after mixing. Since equal volumes are mixed, both solutions are diluted by half: [Pb²⁺] = 0.050 M and [I⁻] = 0.10 M. For the equilibrium $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$, calculate Q using the same expression as K_sp: $Q = [Pb^{2+}][I^-]^2 = (0.050)(0.10)^2 = 5.0 \times 10^{-4}$ Since Q = $5.0 \times 10^{-4}$ and K_sp = $9.8 \times 10^{-9}$, we have Q > K_sp, meaning the solution is supersaturated and a precipitate will form. Choice A is wrong because dilution doesn't prevent precipitation—what matters is whether the diluted concentrations still exceed the solubility limit. Choice C incorrectly states that Q < K_sp, but our calculation shows Q is much larger than K_sp. Choice D mentions the common-ion effect, which describes how adding an ion already present in equilibrium affects solubility, but this isn't the primary reason precipitation occurs here—it's simply that the ion concentrations exceed the solubility limit. Remember: always calculate Q after accounting for dilution effects, then compare to K_sp. If Q > K_sp, precipitation occurs; if Q < K_sp, the solution remains unsaturated.
12. The $K_{sp}$ of $Zn(OH)_2$ is $3.0 \times 10^{-17}$. In a solution where $[OH^-] = 1.0 \times 10^{-4}$ M, what is the equilibrium concentration of $Zn^{2+}$?
    Answer: $3.0 \times 10^{-9}$ M When you encounter solubility equilibrium problems, you need to write the dissolution equation and set up the $K_{sp}$ expression. For zinc hydroxide: $Zn(OH)_2(s) \rightleftharpoons Zn^{2+}(aq) + 2OH^-(aq)$ The $K_{sp}$ expression is: $K_{sp} = [Zn^{2+}][OH^-]^2$ Since you're given both $K_{sp} = 3.0 \times 10^{-17}$ and $[OH^-] = 1.0 \times 10^{-4}$ M, you can solve directly for $[Zn^{2+}]$: $3.0 \times 10^{-17} = [Zn^{2+}](1.0 \times 10^{-4})^2$ $3.0 \times 10^{-17} = [Zn^{2+}](1.0 \times 10^{-8})$ $[Zn^{2+}] = \frac{3.0 \times 10^{-17}}{1.0 \times 10^{-8}} = 3.0 \times 10^{-9}$ M This matches answer A. Looking at the wrong answers: B ($3.0 \times 10^{-13}$ M) results from forgetting to square the hydroxide concentration—using $[OH^-]$ instead of $[OH^-]^2$. C ($1.7 \times 10^{-6}$ M) comes from taking the square root of the correct answer, perhaps confusing this with ICE table problems. D ($5.5 \times 10^{-7}$ M) appears to involve calculation errors in the exponent manipulation. Study tip: Always write the balanced dissolution equation first to get the correct stoichiometry in your $K_{sp}$ expression. The coefficient becomes the exponent—here, the "2" in $Zn(OH)_2$ means $[OH^-]^2$ in the expression. Double-check that you're squaring concentrations when the formula contains subscripts greater than 1.
13. A buffer solution contains 0.25 M $HCOOH$ (formic acid) and 0.35 M $HCOONa$. The $K_a$ for formic acid is $1.8 \times 10^{-4}$. What is the percent ionization of $HCOOH$ in this buffer?
    Answer: 0.051% When you encounter a buffer problem asking for percent ionization, you're dealing with the equilibrium between a weak acid and its conjugate base. The key insight is that buffers significantly suppress the ionization of the weak acid compared to what it would be in pure solution. To find the percent ionization, you first need the $[H^+]$ concentration using the Henderson-Hasselbalch equation or the $K_a$ expression. Since this is a buffer, use: $K_a = \frac{[H^+][A^-]}{[HA]}$ Here, $[A^-] = 0.35 \text{ M}$ (from $HCOONa$) and $[HA] = 0.25 \text{ M}$ ($HCOOH$). Solving: $1.8 \times 10^{-4} = \frac{[H^+](0.35)}{0.25}$ $[H^+] = \frac{1.8 \times 10^{-4} \times 0.25}{0.35} = 1.29 \times 10^{-4} \text{ M}$ Percent ionization = $\frac{[H^+]}{[\text{initial acid}]} \times 100\% = \frac{1.29 \times 10^{-4}}{0.25} \times 100\% = 0.051\%$ This confirms answer C. A) 0.072% likely results from using the wrong concentration ratio or calculation error. B) 0.13% might come from forgetting to account for the buffer effect and treating this like a simple weak acid problem. D) 0.18% could result from incorrectly using the $K_a$ value directly without proper equilibrium calculations. Study tip: In buffer problems, always remember that the conjugate base suppresses ionization dramatically. The percent ionization will be much lower than for the weak acid alone.
14. A solution contains both $Ca^{2+}$ and $Mg^{2+}$ ions, each at 0.010 M concentration. If $Na_2CO_3$ is slowly added to this solution, which carbonate will precipitate first? ($K_{sp}$ values: $CaCO_3 = 3.4 \times 10^{-9}$, $MgCO_3 = 6.8 \times 10^{-6}$)
    Answer: $CaCO_3$ precipitates first because it has a smaller $K_{sp}$ When you encounter selective precipitation problems, the key is determining which compound will reach its solubility limit first as the precipitating agent is gradually added. This happens when the ion product (Q) first equals the solubility product constant ($K_{sp}$). For both carbonates, precipitation begins when $Q = K_{sp} = [M^{2+}][CO_3^{2-}]$, where M represents Ca or Mg. Since both metal ions have identical concentrations (0.010 M), you can solve for the carbonate concentration needed to start each precipitation: For $CaCO_3$: $[CO_3^{2-}] = \frac{K_{sp}}{[Ca^{2+}]} = \frac{3.4 \times 10^{-9}}{0.010} = 3.4 \times 10^{-7}$ M For $MgCO_3$: $[CO_3^{2-}] = \frac{K_{sp}}{[Mg^{2+}]} = \frac{6.8 \times 10^{-6}}{0.010} = 6.8 \times 10^{-4}$ M $CaCO_3$ requires much less carbonate ion to begin precipitating, so it precipitates first. Answer B is correct. Answer A incorrectly assumes larger $K_{sp}$ means earlier precipitation, but the opposite is true—smaller $K_{sp}$ indicates lower solubility. Answer C ignores that different $K_{sp}$ values create different precipitation thresholds despite equal metal ion concentrations. Answer D misapplies the common-ion effect, which isn't relevant here since we're comparing initial precipitation points. Remember: In selective precipitation, calculate the precipitating agent concentration needed for each compound. The one requiring the lowest concentration precipitates first, regardless of which has the larger $K_{sp}$.
15. The equilibrium $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ has $K_{sp} = 1.8 \times 10^{-10}$. If the $[Cl^-]$ in solution is maintained at $1.0 \times 10^{-3}$ M by adding $NaCl$, what is the maximum $[Ag^+]$ that can exist in solution?
    Answer: $1.8 \times 10^{-7}$ M When you encounter a question about solubility equilibrium with a maintained ion concentration, you're dealing with the common ion effect. The key is recognizing that adding NaCl provides excess Cl⁻ ions, which shifts the equilibrium and limits how much AgCl can dissolve. The solubility product expression for this equilibrium is $K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10}$. Since the chloride concentration is maintained at $1.0 \times 10^{-3}$ M by adding NaCl, you can solve directly for the maximum silver ion concentration: $[Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8 \times 10^{-10}}{1.0 \times 10^{-3}} = 1.8 \times 10^{-7} \text{ M}$ This confirms answer A is correct. Looking at the incorrect options: Answer B ($1.8 \times 10^{-10}$ M) represents a common mistake where students confuse the $K_{sp}$ value with the ion concentration. Answer C ($1.3 \times 10^{-5}$ M) might result from calculation errors or incorrect unit conversions. Answer D ($1.8 \times 10^{-13}$ M) could come from mistakenly multiplying the given values instead of dividing, or from mishandling the equilibrium expression. Remember this pattern: when one ion concentration is fixed in a solubility problem, simply rearrange the $K_{sp}$ expression to solve for the unknown ion. The common ion effect always reduces solubility compared to pure water, so expect smaller concentrations than you might see without the added salt.
16. In a solution containing 0.10 M $NH_4^+$ and 0.25 M $NH_3$, the equilibrium $NH_4^+(aq) \rightleftharpoons NH_3(aq) + H^+(aq)$ is established. If the $K_a$ for $NH_4^+$ is $5.6 \times 10^{-10}$, what effect does the common $NH_3$ have on the $[H^+]$?
    Answer: $[H^+]$ decreases due to the common-ion effect When you encounter a question about equilibrium with ions that appear in multiple species, you're dealing with the common-ion effect. This fundamental principle explains how the presence of a shared ion shifts equilibrium position. The correct answer is B because $NH_3$ acts as a common ion that shifts the equilibrium leftward. Using the equilibrium expression $K_a = \frac{[NH_3][H^+]}{[NH_4^+]}$, we can solve for $[H^+]$: $[H^+] = K_a \times \frac{[NH_4^+]}{[NH_3]} = 5.6 \times 10^{-10} \times \frac{0.10}{0.25} = 2.24 \times 10^{-10} M$. If no $NH_3$ were initially present, $[H^+]$ would be much higher. The excess $NH_3$ suppresses the dissociation of $NH_4^+$, reducing $[H^+]$. Choice A incorrectly suggests that more $NH_3$ increases $[H^+]$, but Le Châtelier's principle tells us that adding products shifts equilibrium toward reactants, decreasing $[H^+]$. Choice C misunderstands equilibrium constants—while $K_a$ remains constant, the equilibrium position shifts based on concentrations, affecting $[H^+]$. Choice D incorrectly applies Le Châtelier's principle; adding $NH_3$ (a product) shifts equilibrium left, decreasing $[H^+]$, not increasing it. Remember: whenever you see an equilibrium problem with "common" ions present from the start, think common-ion effect. The shared ion will always suppress the equilibrium, reducing the concentration of products that would form without it.
17. A saturated solution of silver chloride ($AgCl$) has a solubility product constant $K_{sp} = 1.8 \times 10^{-10}$ at 25°C. If 0.10 M $NaCl$ is added to this saturated solution, what happens to the concentration of $Ag^+$ ions?
    Answer: It decreases due to the common-ion effect from $Cl^-$ ions This question tests your understanding of the common-ion effect and solubility equilibrium. When you see a problem involving a saturated solution and the addition of a compound containing a common ion, think about Le Châtelier's principle and how the equilibrium will shift. For $AgCl$, the equilibrium is: $AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}$ with $K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10}$. When you add $NaCl$, it completely dissociates to provide additional $Cl^-$ ions. This increases the $[Cl^-]$ concentration, which means to maintain the constant $K_{sp}$ value, the $[Ag^+]$ concentration must decrease. The equilibrium shifts left, causing some $Ag^+$ ions to combine with the excess $Cl^-$ ions and precipitate as solid $AgCl$. Option A is incorrect because adding salt doesn't automatically increase all ion concentrations—equilibrium considerations matter. Option C shows a fundamental misunderstanding; even though $NaCl$ is different from $AgCl$, they share the common $Cl^-$ ion, which directly affects the equilibrium. Option D incorrectly suggests that ionic strength effects would dominate over the common-ion effect, but the common-ion effect is the primary factor here. Remember: whenever you add a compound containing an ion that's already present in an equilibrium system, expect the common-ion effect to shift the equilibrium away from producing more of that ion. This is a key concept in solubility problems.
18. Consider the equilibrium: $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)$. A solution at equilibrium appears red due to the $FeSCN^{2+}$ complex. If solid $KSCN$ is added to this solution, what will be observed?
    Answer: The solution becomes more red due to increased $FeSCN^{2+}$ formation When you encounter equilibrium problems involving color changes, think about Le Châtelier's principle: how will the system respond to restore equilibrium when you disturb it? Adding solid $KSCN$ increases the concentration of $SCN^-$ ions in solution. According to Le Châtelier's principle, the equilibrium will shift to counteract this disturbance by consuming the excess $SCN^-$. The reaction shifts right, forming more $FeSCN^{2+}$ complex. Since $FeSCN^{2+}$ is responsible for the red color, increasing its concentration makes the solution more intensely red. This confirms answer B. Let's examine why the other options are incorrect. Option A suggests dilution causes the color change, but adding solid $KSCN$ doesn't significantly change the solution volume, and the chemical effect (Le Châtelier's principle) dominates anyway. Option C incorrectly assumes $KSCN$ won't affect equilibrium, but $KSCN$ dissociates to provide $SCN^-$, which is directly involved in the equilibrium reaction. Option D wrongly suggests the solution becomes colorless by removing $Fe^{3+}$, but $SCN^-$ doesn't remove iron ions—it combines with them to form more colored complex. Remember this pattern: when you add a reactant to an equilibrium system, the reaction shifts toward the products to consume the excess reactant. Always identify which species you're adding and trace how the equilibrium responds to restore balance.
19. A solution is prepared by mixing 50.0 mL of 0.20 M $CH_3COOH$ with 30.0 mL of 0.15 M $CH_3COONa$. What is the $[CH_3COOH]$ in the final solution?
    Answer: 0.125 M When you're mixing two solutions, you need to find the final concentration by considering both the moles of each substance and the new total volume. This is a dilution problem combined with solution mixing. First, calculate the moles of $CH_3COOH$: (0.0500 L)(0.20 M) = 0.0100 mol. The $CH_3COONa$ doesn't affect the amount of acetic acid present—it's a separate compound that will create a buffer system, but the acetic acid molecules themselves remain unchanged. Next, find the total final volume: 50.0 mL + 30.0 mL = 80.0 mL = 0.0800 L. Now you can calculate the final concentration of $CH_3COOH$: $\frac{0.0100 \text{ mol}}{0.0800 \text{ L}} = 0.125 \text{ M}$ Looking at the wrong answers: B) 0.113 M likely comes from an error in volume conversion or arithmetic. C) 0.200 M is the original concentration of the acetic acid—this would be wrong because it ignores the dilution effect of adding the sodium acetate solution. D) 0.175 M might result from incorrectly averaging the two original concentrations without considering the different volumes. The key insight is that A) 0.125 M correctly accounts for the simple dilution of acetic acid when the sodium acetate solution is added. Study tip: In solution mixing problems, always track moles of each species separately, then divide by the total final volume. Don't be distracted by the chemical relationship between the compounds—focus on the math of dilution first.
20. A solution contains 0.040 M $H_2S$ and 0.15 M $HS^-$. Given that $K_{a1}$ for $H_2S$ is $9.5 \times 10^{-8}$, what is the $[H^+]$ in this solution?
    Answer: $2.5 \times 10^{-8}$ M When you encounter a problem with both a weak acid and its conjugate base present in solution, you're dealing with a buffer system. The key insight is to use the Henderson-Hasselbalch equation or directly apply the acid dissociation expression. For the first dissociation of $H_2S$: $H_2S \rightleftharpoons H^+ + HS^-$ The equilibrium expression is: $K_{a1} = \frac{[H^+][HS^-]}{[H_2S]}$ Since you're given concentrations of both $H_2S$ (0.040 M) and $HS^-$ (0.15 M), you can solve directly for $[H^+]$: $9.5 \times 10^{-8} = \frac{[H^+](0.15)}{(0.040)}$ $[H^+] = \frac{(9.5 \times 10^{-8})(0.040)}{0.15} = 2.5 \times 10^{-8} \text{ M}$ This confirms answer A is correct. Answer B ($9.5 \times 10^{-8}$ M) represents the common mistake of confusing $[H^+]$ with $K_{a1}$ itself. Answer C ($1.3 \times 10^{-7}$ M) likely comes from incorrectly flipping the concentration ratio in the calculation. Answer D ($3.6 \times 10^{-8}$ M) appears to result from arithmetic errors in the setup. Remember: in buffer problems, the $[H^+]$ depends on both the $K_a$ value and the ratio of conjugate base to acid concentrations. Always set up the equilibrium expression carefully and substitute the given concentrations directly.

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