Introduction to Solubility Equilibria - College Chemistry

1. The $K_{sp}$ for calcium fluoride ($CaF_2$) is $3.9 \times 10^{-11}$ at 25°C. What is the molar solubility of $CaF_2$ in pure water at this temperature?
   Answer: $2.1 \times 10^{-4}$ M When you encounter a solubility product ($K_{sp}$) problem, you're dealing with equilibrium expressions for slightly soluble ionic compounds. The key is setting up the dissolution equation and relating the concentrations of ions to the molar solubility. For calcium fluoride, the dissolution equilibrium is: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$ Let $s$ represent the molar solubility of $CaF_2$. When $s$ moles of $CaF_2$ dissolve per liter, you get $s$ moles of $Ca^{2+}$ ions and $2s$ moles of $F^-$ ions (notice the 1:2 stoichiometry from the formula). The $K_{sp}$ expression becomes: $K_{sp} = [Ca^{2+}][F^-]^2 = (s)(2s)^2 = 4s^3$ Substituting the given $K_{sp}$ value: $3.9 \times 10^{-11} = 4s^3$ Solving for $s$: $s^3 = \frac{3.9 \times 10^{-11}}{4} = 9.75 \times 10^{-12}$ Therefore: $s = \sqrt[3]{9.75 \times 10^{-12}} = 2.1 \times 10^{-4}$ M, which is answer B. Answer A ($1.3 \times 10^{-11}$ M) appears to confuse $K_{sp}$ with solubility directly. Answer C ($6.2 \times 10^{-6}$ M) likely results from incorrectly using $K_{sp} = s^2$ instead of $4s^3$. Answer D ($1.2 \times 10^{-5}$ M) may come from calculation errors in the cube root step. Remember: always write the balanced dissolution equation first, then express each ion concentration in terms of molar solubility, accounting for stoichiometric coefficients.
2. A solution contains $0.010$ M $Ca^{2+}$ and $0.020$ M $SO_4^{2-}$. The $K_{sp}$ for $CaSO_4$ is $2.4 \times 10^{-5}$. What will happen when these solutions are mixed?
   Answer: Precipitation occurs because $Q > K_{sp}$ When you encounter a precipitation problem, you need to compare the reaction quotient (Q) to the solubility product constant ($K_{sp}$) to predict whether a precipitate will form. First, calculate the reaction quotient using the given concentrations. For the equilibrium $CaSO_4(s) \rightleftharpoons Ca^{2+}(aq) + SO_4^{2-}(aq)$, the expression is $Q = [Ca^{2+}][SO_4^{2-}]$. Substituting the values: $Q = (0.010)(0.020) = 2.0 \times 10^{-4}$ Now compare Q to $K_{sp}$: $Q = 2.0 \times 10^{-4}$ versus $K_{sp} = 2.4 \times 10^{-5}$ Since $2.0 \times 10^{-4} > 2.4 \times 10^{-5}$, we have $Q > K_{sp}$, which means the solution is supersaturated and precipitation will occur. Looking at the wrong answers: Choice A incorrectly states that $Q < K_{sp}$, which would mean no precipitation occurs, but our calculation shows Q is actually larger than $K_{sp}$. Choice B suggests $Q = K_{sp}$, which would indicate equilibrium with no net precipitation, but Q and $K_{sp}$ are clearly different values. Choice D implies the solution reaches exact saturation, which only occurs when $Q = K_{sp}$, not when Q exceeds $K_{sp}$. Study tip: Always calculate Q first, then compare it to $K_{sp}$. Remember the rule: Q > $K_{sp}$ means precipitation occurs, Q < $K_{sp}$ means no precipitation, and Q = $K_{sp}$ means the solution is at equilibrium (saturated).
3. The $K_{sp}$ values for several lead compounds are given below. Which compound has the highest molar solubility in pure water?
   Answer: $PbCl_2$ ($K_{sp} = 1.6 \times 10^{-5}$) When you encounter questions comparing molar solubilities using $K_{sp}$ values, you can't simply compare the $K_{sp}$ numbers directly. The key is understanding that compounds with different stoichiometries require different mathematical relationships between $K_{sp}$ and molar solubility. For compounds with the formula $MX_2$ (like $PbCl_2$, $PbBr_2$, and $PbI_2$), the dissolution equation is $MX_2 \rightleftharpoons M^{2+} + 2X^-$. If the molar solubility is $s$, then $[M^{2+}] = s$ and $[X^-] = 2s$, giving $K_{sp} = s \times (2s)^2 = 4s^3$. Therefore, $s = \sqrt[3]{\frac{K_{sp}}{4}}$. For $PbS$ with formula $MX$, the equation is $MX \rightleftharpoons M^{2+} + X^{2-}$, so $K_{sp} = s^2$ and $s = \sqrt{K_{sp}}$. Calculating the molar solubilities: - **A) $PbCl_2$**: $s = \sqrt[3]{\frac{1.6 \times 10^{-5}}{4}} = 1.6 \times 10^{-2}$ M - **B) $PbBr_2$**: $s = \sqrt[3]{\frac{4.0 \times 10^{-6}}{4}} = 1.0 \times 10^{-2}$ M - **C) $PbI_2$**: $s = \sqrt[3]{\frac{7.1 \times 10^{-9}}{4}} = 1.2 \times 10^{-3}$ M - **D) $PbS$**: $s = \sqrt{3.0 \times 10^{-28}} = 5.5 \times 10^{-15}$ M $PbCl_2$ has the highest molar solubility, making A correct. Study tip: Always write the dissolution equation first to determine the relationship between $K_{sp}$ and molar solubility. Different stoichiometries require different formulas, so comparing $K_{sp}$ values alone can be misleading.
4. A student measures the conductivity of a saturated $AgCl$ solution and finds it to be very low. Which of the following best explains this observation?
   Answer: The $K_{sp}$ of $AgCl$ is very small, resulting in few ions in solution When you encounter questions about conductivity and ionic compounds, focus on the relationship between ion concentration and electrical conductivity. Conductivity depends on the number of ions present in solution - more ions mean better conductivity. Silver chloride ($AgCl$) is an ionic compound that dissociates according to: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$. However, $AgCl$ has an extremely small solubility product constant ($K_{sp} = 1.8 \times 10^{-10}$), meaning very little dissolves. In a saturated solution, only tiny amounts of $Ag^+$ and $Cl^-$ ions are present, resulting in very low conductivity. This explains why option B is correct. Option A is incorrect because $AgCl$ is definitely an ionic compound (formed between a metal and nonmetal with a large electronegativity difference), and it does ionize - just very sparingly. Option C misidentifies the issue; while ion pairing can occur in some solutions, the primary reason for low conductivity here is simply the extremely low concentration of ions due to poor solubility. Option D contradicts the scenario - you can't have a "too concentrated" saturated solution, and higher ion concentrations would actually increase conductivity, not decrease it. Remember this pattern: when you see low conductivity paired with an ionic compound, immediately think about solubility and $K_{sp}$ values. Sparingly soluble salts like $AgCl$, $BaSO_4$, and $PbI_2$ will always show low conductivity because they produce few ions in solution.
5. A saturated solution of silver chloride ($AgCl$) at 25°C has a concentration of $Ag^+$ ions equal to $1.3 \times 10^{-5}$ M. What is the solubility product constant ($K_{sp}$) for $AgCl$ at this temperature?
   Answer: $1.7 \times 10^{-10}$ When you encounter solubility product problems, you're dealing with equilibrium expressions for slightly soluble ionic compounds. The key is understanding how the dissolution equation relates to the $K_{sp}$ expression. Silver chloride dissolves according to: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ The solubility product expression is: $K_{sp} = [Ag^+][Cl^-]$ Since the stoichiometry is 1:1, each mole of $AgCl$ that dissolves produces one mole of $Ag^+$ and one mole of $Cl^-$. Therefore, if $[Ag^+] = 1.3 \times 10^{-5}$ M, then $[Cl^-] = 1.3 \times 10^{-5}$ M as well. Substituting into the $K_{sp}$ expression: $K_{sp} = (1.3 \times 10^{-5})(1.3 \times 10^{-5}) = 1.69 \times 10^{-10}$ This rounds to $1.7 \times 10^{-10}$, which is answer B. Answer A ($1.3 \times 10^{-5}$) represents just the concentration of $Ag^+$, not the product. Answer C ($2.6 \times 10^{-5}$) is the sum of the ion concentrations, which isn't relevant to $K_{sp}$. Answer D ($6.5 \times 10^{-6}$) might result from incorrectly dividing instead of multiplying the concentrations. Remember: For $K_{sp}$ calculations, always write the balanced dissolution equation first, then the equilibrium expression. Use stoichiometry to find all ion concentrations, then multiply them together according to the $K_{sp}$ expression.
6. Which of the following best explains the student's observation that $Mg(OH)_2$ is more soluble at lower pH?
   Answer: $H^+$ ions react with $OH^-$ ions, reducing $[OH^-]$ and allowing more $Mg(OH)_2$ to dissolve When you encounter questions about pH affecting the solubility of metal hydroxides, think about Le Châtelier's principle and how changing conditions shift equilibrium positions. The dissolution of $Mg(OH)_2$ establishes this equilibrium: $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$. When you lower the pH by adding acid, you're increasing the concentration of $H^+$ ions. These $H^+$ ions react with the $OH^-$ ions in solution: $H^+ + OH^- \rightarrow H_2O$. This reaction removes $OH^-$ ions from the equilibrium, decreasing $[OH^-]$. According to Le Châtelier's principle, the system responds by shifting right to replace the consumed $OH^-$ ions, dissolving more $Mg(OH)_2$ in the process. Option A is incorrect because $K_{sp}$ is a constant at a given temperature—it doesn't change with pH. The equilibrium expression $K_{sp} = [Mg^{2+}][OH^-]^2$ remains the same, but the individual ion concentrations adjust. Option C misses the specific mechanism. While acidic solutions do contain more ions, this doesn't directly explain why $Mg(OH)_2$ becomes more soluble—it's specifically the consumption of $OH^-$ ions that matters. Option D suggests $Mg^{2+}$ stability changes with pH, but this isn't the driving force. The magnesium ion itself isn't significantly affected by pH changes in this context. Remember: When hydroxide salts encounter acidic conditions, look for the $H^+ + OH^- \rightarrow H_2O$ reaction. This "removes" one of the products from the dissolution equilibrium, driving more solid to dissolve.
7. The $K_{sp}$ for $Cu(OH)_2$ is $2.2 \times 10^{-20}$ at 25°C. In a solution where $[OH^-] = 1.0 \times 10^{-5}$ M, what is the maximum concentration of $Cu^{2+}$ that can exist without precipitation?
   Answer: $2.2 \times 10^{-10}$ M This question tests your understanding of solubility equilibrium and how to use $K_{sp}$ expressions to find maximum ion concentrations before precipitation occurs. When $Cu(OH)_2$ dissolves, it establishes the equilibrium: $Cu(OH)_2(s) \rightleftharpoons Cu^{2+}(aq) + 2OH^-(aq)$. The $K_{sp}$ expression is: $K_{sp} = [Cu^{2+}][OH^-]^2$. Notice that $[OH^-]$ is squared because there are two hydroxide ions per formula unit. To find the maximum $[Cu^{2+}]$ before precipitation, substitute the known values into the $K_{sp}$ expression: $2.2 \times 10^{-20} = [Cu^{2+}](1.0 \times 10^{-5})^2$. Solving for $[Cu^{2+}]$: $[Cu^{2+}] = \frac{2.2 \times 10^{-20}}{(1.0 \times 10^{-5})^2} = \frac{2.2 \times 10^{-20}}{1.0 \times 10^{-10}} = 2.2 \times 10^{-10}$ M. Answer A ($2.2 \times 10^{-10}$ M) is correct. Answer B ($2.2 \times 10^{-15}$ M) results from incorrectly multiplying instead of dividing by $[OH^-]^2$. Answer C ($4.7 \times 10^{-8}$ M) comes from forgetting to square the $[OH^-]$ term, using $(1.0 \times 10^{-5})^1$ instead of $(1.0 \times 10^{-5})^2$. Answer D ($1.5 \times 10^{-7}$ M) appears to involve calculation errors or incorrect substitution. Remember: always check the stoichiometric coefficients in the balanced equation to write the correct $K_{sp}$ expression. The exponents in $K_{sp}$ expressions must match the coefficients from the dissolution equation.
8. The $K_{sp}$ for $Mg(OH)_2$ is $1.8 \times 10^{-11}$ at 25°C. What is the pH of a saturated solution of $Mg(OH)_2$ in pure water?
   Answer: 10.5 This question tests your ability to connect solubility equilibria with acid-base chemistry. When a sparingly soluble hydroxide dissolves, it releases OH⁻ ions that affect the solution's pH. Start by writing the dissolution equilibrium: $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$. The $K_{sp}$ expression is $K_{sp} = [Mg^{2+}][OH^-]^2 = 1.8 \times 10^{-11}$. Let the molar solubility of $Mg(OH)_2$ be $s$. At equilibrium, $[Mg^{2+}] = s$ and $[OH^-] = 2s$ (note the 2:1 stoichiometry). Substituting: $K_{sp} = (s)(2s)^2 = 4s^3 = 1.8 \times 10^{-11}$. Solving for $s$: $s^3 = \frac{1.8 \times 10^{-11}}{4} = 4.5 \times 10^{-12}$, so $s = 1.65 \times 10^{-4}$ M. Therefore, $[OH^-] = 2s = 3.3 \times 10^{-4}$ M. Calculate pOH: $pOH = -\log(3.3 \times 10^{-4}) = 3.5$. Finally, $pH = 14 - pOH = 14 - 3.5 = 10.5$. Choice A (9.8) likely results from forgetting the factor of 2 in the hydroxide concentration. Choice B (10.2) might come from calculation errors in the cube root step. Choice D (11.1) could result from incorrectly assuming $[OH^-] = s$ instead of $2s$, then making additional errors. Remember: always pay attention to stoichiometry when writing equilibrium expressions. The coefficient in the balanced equation becomes an exponent in $K_{sp}$ and a multiplier for ion concentrations.
9. A saturated solution of $CaCO_3$ is in equilibrium with solid $CaCO_3$. If $HCl$ is added to this solution, which of the following will occur?
   Answer: More $CaCO_3$ will dissolve due to consumption of $CO_3^{2-}$ ions This question tests your understanding of how acid-base reactions affect solubility equilibria. When you encounter a problem involving a saturated solution and added acid, think about how the acid will react with the ions in solution and shift the equilibrium. The saturated $CaCO_3$ solution exists in equilibrium: $CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$. When $HCl$ is added, the $H^+$ ions react with carbonate ions to form bicarbonate and eventually carbonic acid: $CO_3^{2-} + H^+ \rightarrow HCO_3^- + H^+ \rightarrow H_2CO_3$. This removes $CO_3^{2-}$ ions from solution, which shifts the equilibrium to the right according to Le Châtelier's principle, causing more solid $CaCO_3$ to dissolve. This confirms answer C. Answer A incorrectly suggests precipitation due to a common ion effect, but $HCl$ doesn't provide any ions common to the $CaCO_3$ equilibrium. Answer B is wrong because $K_{sp}$ is a constant that depends only on temperature, not on the presence of other substances. The acid doesn't change the fundamental solubility constant. Answer D suggests supersaturation, but removing carbonate ions actually decreases the ion product below $K_{sp}$, making the solution undersaturated. Remember: when acids are added to solutions containing basic anions like $CO_3^{2-}$, $OH^-$, or $S^{2-}$, they consume these anions through acid-base reactions, which increases the solubility of the corresponding salts by shifting equilibrium toward dissolution.
10. Which of the following statements about solubility equilibria is correct?
    Answer: The $K_{sp}$ expression includes only the concentrations of dissolved ions Solubility equilibria questions test your understanding of how ionic compounds dissolve and reach equilibrium in solution. When a slightly soluble ionic compound dissolves, it establishes an equilibrium between the solid and its dissolved ions. The solubility product constant ($K_{sp}$) expression follows the same rules as other equilibrium expressions, but with one crucial difference: pure solids are never included in equilibrium expressions because their concentration remains constant. For example, when $\text{AgCl}$ dissolves: $\text{AgCl}_{(s)} \rightleftharpoons \text{Ag}^+_{(aq)} + \text{Cl}^-_{(aq)}$, the $K_{sp}$ expression is $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$. Only the concentrations of the dissolved ions appear in the expression, making option C correct. Option A is wrong because, as mentioned, pure solids never appear in equilibrium expressions since their "concentration" doesn't change during the reaction. Option B contains a common misconception: while larger $K_{sp}$ values often indicate greater solubility, you can't directly compare $K_{sp}$ values between compounds that produce different numbers of ions. For instance, $\text{Mg(OH)}_2$ produces three ions while $\text{AgCl}$ produces two, so their $K_{sp}$ values aren't directly comparable for solubility. Option D is incorrect because $K_{sp}$ values are quite temperature-dependent, just like other equilibrium constants. Remember: $K_{sp}$ expressions only include aqueous ions, never the solid reactant. When comparing solubilities, calculate actual molar solubilities rather than just comparing $K_{sp}$ values.
11. A solution contains $0.015$ M $Pb^{2+}$ ions. What minimum concentration of $Cl^-$ ions is required to initiate precipitation of $PbCl_2$? The $K_{sp}$ for $PbCl_2$ is $1.6 \times 10^{-5}$.
    Answer: $3.3 \times 10^{-2}$ M When you encounter a precipitation problem, you're dealing with solubility equilibrium and the solubility product constant ($K_{sp}$). Precipitation begins when the ion product equals $K_{sp}$, meaning the solution becomes saturated. For $PbCl_2$, the dissolution equilibrium is: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$ The $K_{sp}$ expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2 = 1.6 \times 10^{-5}$ Notice that the chloride concentration is squared because there are two $Cl^-$ ions per formula unit of $PbCl_2$. To find the minimum $[Cl^-]$ needed for precipitation, substitute the given $[Pb^{2+}] = 0.015$ M into the $K_{sp}$ expression: $1.6 \times 10^{-5} = (0.015)[Cl^-]^2$ Solving for $[Cl^-]$: $[Cl^-]^2 = \frac{1.6 \times 10^{-5}}{0.015} = 1.07 \times 10^{-3}$ $[Cl^-] = \sqrt{1.07 \times 10^{-3}} = 3.3 \times 10^{-2}$ M This matches answer choice B. Answer A ($1.1 \times 10^{-6}$ M) likely comes from incorrectly dividing $K_{sp}$ by $[Pb^{2+}]$ without taking the square root. Answer C ($1.1 \times 10^{-3}$ M) represents the intermediate step before taking the square root. Answer D ($4.0 \times 10^{-4}$ M) might result from calculation errors or forgetting to account for the stoichiometry. Remember: always write the correct $K_{sp}$ expression first, paying careful attention to stoichiometric coefficients that become exponents.
12. Which of the following experimental observations would provide the strongest evidence that a solution is saturated with respect to $BaSO_4$?
    Answer: Adding more solid $BaSO_4$ does not increase the concentration of dissolved ions When you encounter questions about solution saturation, focus on the defining characteristic: a saturated solution exists in dynamic equilibrium with its undissolved solute. At this point, the rates of dissolution and precipitation are equal, meaning no net change occurs in ion concentration. The strongest evidence for saturation is when adding more solid $BaSO_4$ fails to increase the concentration of dissolved ions (B). This directly demonstrates that the solution has reached its maximum capacity to hold dissolved $Ba^{2+}$ and $SO_4^{2-}$ ions. Any additional solid will simply remain undissolved, maintaining the equilibrium. Option A is misleading because $BaSO_4$ has extremely low solubility regardless of saturation status - even an unsaturated solution would show minimal conductivity compared to pure water. Option C describes a supersaturated or unstable solution with excess solid, but cloudiness from suspended particles doesn't prove the dissolved portion is at saturation - you could have undissolved solid in an unsaturated solution due to slow dissolution kinetics. Option D is irrelevant since $BaSO_4$ is a salt of a strong acid and strong base, so it won't significantly alter pH whether saturated or not. Remember this key principle: saturation is specifically about the dissolved species being at maximum concentration. Look for evidence that adding more solute fails to increase the amount that dissolves - this is the definitive test for a saturated solution and distinguishes it from simply having undissolved solid present.
13. Consider the equilibrium: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$. Which of the following changes would increase the solubility of $CaF_2$?
    Answer: Adding $HCl$ to the solution When you encounter equilibrium problems involving ionic solids, think about Le Châtelier's principle and how changes affect the position of equilibrium. The key is identifying which changes will shift the equilibrium toward more dissolution (right) versus precipitation (left). Adding $HCl$ increases $CaF_2$ solubility because the added $H^+$ ions react with $F^-$ ions to form $HF$, effectively removing $F^-$ from solution. According to Le Châtelier's principle, when you remove a product ($F^-$), the equilibrium shifts right to replace it, dissolving more $CaF_2$. This is called the common ion effect in reverse—removing an ion rather than adding one. Choice A is wrong because adding $NaF$ increases the $F^-$ concentration, creating a common ion effect that shifts equilibrium left, decreasing solubility. Choice B is incorrect because adding $CaCl_2$ increases $Ca^{2+}$ concentration, another common ion effect that also shifts equilibrium left and decreases solubility. Choice D is wrong because if dissolution is exothermic (releases heat), increasing temperature shifts equilibrium left toward the solid, decreasing solubility. Remember this pattern: to increase solubility of an ionic compound, you need to remove one of its ions from solution (through chemical reaction) or add something that doesn't share common ions. Common ion effects always decrease solubility, while reactions that consume the dissolved ions increase it.
14. A solution is prepared by dissolving $0.025$ mol of $KI$ in enough water to make $500$ mL of solution. If $Pb(NO_3)_2$ is slowly added to this solution, at what $[Pb^{2+}]$ will $PbI_2$ begin to precipitate? The $K_{sp}$ for $PbI_2$ is $7.1 \times 10^{-9}$.
    Answer: $2.8 \times 10^{-6}$ M When you encounter precipitation problems, you're working with solubility equilibria and the concept that precipitation begins when the ion product equals the solubility product constant ($K_{sp}$). First, find the iodide ion concentration. You have 0.025 mol of KI in 500 mL (0.500 L) of solution. Since KI dissociates completely: $[I^-] = \frac{0.025 \text{ mol}}{0.500 \text{ L}} = 0.050 \text{ M}$ For $PbI_2$ precipitation, the equilibrium expression is: $K_{sp} = [Pb^{2+}][I^-]^2 = 7.1 \times 10^{-9}$ Notice that the iodide concentration is squared because the formula shows two iodide ions per lead ion. Precipitation begins when this expression equals $K_{sp}$: $[Pb^{2+}] = \frac{K_{sp}}{[I^-]^2} = \frac{7.1 \times 10^{-9}}{(0.050)^2} = \frac{7.1 \times 10^{-9}}{2.5 \times 10^{-3}} = 2.8 \times 10^{-6} \text{ M}$ This confirms answer B. Looking at the distractors: A ($7.1 \times 10^{-6}$ M) appears to result from not squaring the iodide concentration. C ($1.4 \times 10^{-7}$ M) might come from incorrectly using the molarity as 0.1 M instead of 0.05 M. D ($3.6 \times 10^{-8}$ M) could result from multiple calculation errors or using an incorrect volume conversion. Remember: always write the correct equilibrium expression with proper stoichiometry, and precipitation begins exactly when $Q = K_{sp}$, not when it exceeds it.
15. Which of the following ionic compounds would be expected to have the lowest molar solubility in pure water, based on the given $K_{sp}$ values?
    Answer: $AgBr$ ($K_{sp} = 5.4 \times 10^{-13}$) When comparing ionic compounds' solubilities, you can't simply compare $K_{sp}$ values directly—you must consider how each compound dissociates and calculate the actual molar solubility. For each compound, write the dissolution equation and express $K_{sp}$ in terms of molar solubility (s): AgBr: $AgBr \rightleftharpoons Ag^+ + Br^-$ $K_{sp} = [Ag^+][Br^-] = s \cdot s = s^2$ $s = \sqrt{5.4 \times 10^{-13}} = 7.3 \times 10^{-7} \text{ M}$ CaF₂: $CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$ $K_{sp} = [Ca^{2+}][F^-]^2 = s \cdot (2s)^2 = 4s^3$ $s = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}} = 2.1 \times 10^{-4} \text{ M}$ Ag₂SO₄: $Ag_2SO_4 \rightleftharpoons 2Ag^+ + SO_4^{2-}$ $K_{sp} = [Ag^+]^2[SO_4^{2-}] = (2s)^2 \cdot s = 4s^3$ $s = \sqrt[3]{\frac{1.2 \times 10^{-5}}{4}} = 1.4 \times 10^{-2} \text{ M}$ PbI₂: $PbI_2 \rightleftharpoons Pb^{2+} + 2I^-$ $K_{sp} = [Pb^{2+}][I^-]^2 = s \cdot (2s)^2 = 4s^3$ $s = \sqrt[3]{\frac{7.1 \times 10^{-9}}{4}} = 1.2 \times 10^{-3} \text{ M}$ Choice A (AgBr) has the lowest molar solubility at $7.3 \times 10^{-7}$ M. Choices B, C, and D all have significantly higher solubilities because their stoichiometry creates multiple ions, making them more soluble despite having larger $K_{sp}$ values in some cases. Key strategy: Always convert $K_{sp}$ to molar solubility when comparing compounds with different stoichiometries. A smaller $K_{sp}$ doesn't automatically mean lower solubility if the dissolution produces different numbers of ions.
16. The molar solubility of $Ag_2CrO_4$ in pure water is $6.5 \times 10^{-5}$ M at 25°C. What is the $K_{sp}$ for $Ag_2CrO_4$?
    Answer: $1.1 \times 10^{-12}$ When you encounter solubility equilibrium problems, you need to connect molar solubility (how much dissolves) to the solubility product constant ($K_{sp}$) through the dissolution equation and stoichiometry. For $Ag_2CrO_4$, the dissolution equilibrium is: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$ The key insight is understanding the stoichiometric relationship. If the molar solubility is $6.5 \times 10^{-5}$ M, this means $6.5 \times 10^{-5}$ mol of $Ag_2CrO_4$ dissolves per liter. Since each formula unit produces 2 $Ag^+$ ions and 1 $CrO_4^{2-}$ ion: - $[Ag^+] = 2 \times 6.5 \times 10^{-5} = 1.3 \times 10^{-4}$ M - $[CrO_4^{2-}] = 6.5 \times 10^{-5}$ M The $K_{sp}$ expression is: $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$ Substituting: $K_{sp} = (1.3 \times 10^{-4})^2 \times (6.5 \times 10^{-5}) = 1.69 \times 10^{-8} \times 6.5 \times 10^{-5} = 1.1 \times 10^{-12}$ This matches answer B. Answer A ($2.7 \times 10^{-13}$) likely comes from incorrectly squaring the entire molar solubility. Answer C ($4.2 \times 10^{-9}$) suggests forgetting to account for the stoichiometric coefficient of 2 for $Ag^+$. Answer D ($2.8 \times 10^{-8}$) appears to involve calculation errors in the exponent arithmetic. Remember: always write the balanced dissolution equation first, then use stoichiometry to find individual ion concentrations before calculating $K_{sp}$.
17. The $K_{sp}$ for $BaSO_4$ is $1.1 \times 10^{-10}$ at 25°C. What is the solubility of $BaSO_4$ in a $0.050$ M $Na_2SO_4$ solution?
    Answer: $2.2 \times 10^{-9}$ M This question tests the common ion effect, where the presence of a shared ion reduces the solubility of a sparingly soluble salt. When you see a solubility problem with an added salt that shares an ion, expect significantly reduced solubility compared to pure water. For $BaSO_4$, the equilibrium is: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$ with $K_{sp} = [Ba^{2+}][SO_4^{2-}] = 1.1 \times 10^{-10}$. The $Na_2SO_4$ completely dissociates, providing 0.050 M $SO_4^{2-}$ ions. When $BaSO_4$ dissolves, it adds $x$ M of both $Ba^{2+}$ and $SO_4^{2-}$. Since $x$ will be very small compared to 0.050 M, the total sulfate concentration is approximately 0.050 M. Setting up the $K_{sp}$ expression: $1.1 \times 10^{-10} = (x)(0.050)$ Solving for $x$: $x = \frac{1.1 \times 10^{-10}}{0.050} = 2.2 \times 10^{-9}$ M This confirms answer A is correct. Answer B ($1.0 \times 10^{-5}$ M) might result from incorrectly using the square root of $K_{sp}$, forgetting about the common ion effect. Answer C ($5.0 \times 10^{-6}$ M) could come from calculation errors or wrong assumptions about ion concentrations. Answer D ($1.1 \times 10^{-8}$ M) might arise from incorrectly manipulating the $K_{sp}$ expression. Remember: when a common ion is present, always account for its initial concentration before adding the dissolving salt's contribution. The common ion effect dramatically reduces solubility.
18. Based on the scenario described above, which of the following anions would produce the least amount of $AgX$ precipitate when mixed with $AgNO_3$?
    Answer: $NO_3^-$ ($AgNO_3$ is highly soluble) When you encounter precipitation questions, focus on two key factors: whether a precipitate can form at all, and if so, how much forms based on solubility equilibrium. The correct answer is D because $AgNO_3$ is highly soluble in water. When you mix $NO_3^-$ ions with $Ag^+$ ions, no precipitate forms whatsoever since silver nitrate remains completely dissolved. This means zero precipitate formation, which is definitively the least amount possible. Let's examine why the other options are wrong. Choice A ($Cl^-$) will form $AgCl$ precipitate with $K_{sp} = 1.8 \times 10^{-10}$. Choice B ($Br^-$) forms $AgBr$ with $K_{sp} = 5.4 \times 10^{-13}$. Choice C ($I^-$) produces $AgI$ with $K_{sp} = 8.3 \times 10^{-17}$. All three of these will form precipitates because their $K_{sp}$ values are extremely small, indicating very low solubility. While $AgCl$ would form the least precipitate among these three (since it has the largest $K_{sp}$), it still forms significantly more precipitate than the zero amount produced with $NO_3^-$. Remember this strategy: when comparing precipitation amounts, first determine which combinations actually form precipitates. A highly soluble compound that forms no precipitate will always produce less solid than even the most soluble precipitate-forming combination. Don't get distracted by comparing $K_{sp}$ values if one option doesn't precipitate at all.
19. The $K_{sp}$ for $PbCl_2$ is $1.6 \times 10^{-5}$ at 25°C. What is the molar solubility of $PbCl_2$ in a $0.10$ M $NaCl$ solution?
    Answer: $1.6 \times 10^{-3}$ M This question tests your understanding of the common ion effect in solubility equilibria. When a sparingly soluble salt dissolves in a solution containing one of its ions, the solubility decreases due to Le Châtelier's principle. For $PbCl_2$, the equilibrium is: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$ The $K_{sp}$ expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2 = 1.6 \times 10^{-5}$ In the $0.10$ M $NaCl$ solution, the initial $[Cl^-] = 0.10$ M. Let $s$ be the molar solubility of $PbCl_2$. At equilibrium: - $[Pb^{2+}] = s$ - $[Cl^-] = 0.10 + 2s$ Since $s$ will be small compared to $0.10$ M, we can approximate: $[Cl^-] \approx 0.10$ M Substituting into the $K_{sp}$ expression: $1.6 \times 10^{-5} = s \times (0.10)^2$ $s = \frac{1.6 \times 10^{-5}}{0.01} = 1.6 \times 10^{-3}$ M Answer A ($1.6 \times 10^{-3}$ M) is correct. Answer B ($4.0 \times 10^{-4}$ M) might result from calculation errors or incorrect stoichiometry. Answer C ($1.6 \times 10^{-4}$ M) could come from using $0.10$ instead of $(0.10)^2$ in the denominator. Answer D ($1.6 \times 10^{-5}$ M) incorrectly assumes the molar solubility equals the $K_{sp}$ value. Remember: when dealing with common ion effects, identify which ion is common, set up your ICE table carefully, and don't forget to square the chloride concentration in the $K_{sp}$ expression.
20. For the dissolution reaction $CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$, which of the following correctly represents the relationship between the molar solubility ($s$) and the equilibrium concentrations?
    Answer: $[Ca^{2+}] = s$ and $[CO_3^{2-}] = s$ When you encounter solubility equilibrium problems, the key is understanding that molar solubility represents the amount of solid compound that dissolves per liter of solution. You need to connect this dissolution amount to the concentrations of individual ions produced. For the dissolution $CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$, examine the stoichiometry carefully. When one mole of $CaCO_3$ dissolves, it produces exactly one mole of $Ca^{2+}$ ions and one mole of $CO_3^{2-}$ ions. If the molar solubility is $s$ (meaning $s$ moles of $CaCO_3$ dissolve per liter), then $s$ moles of each ion are produced per liter. Therefore, $[Ca^{2+}] = s$ and $[CO_3^{2-}] = s$, making choice A correct. Choice B incorrectly suggests $[Ca^{2+}] = 2s$, which would only be true if each $CaCO_3$ unit produced two calcium ions (like $Ca_2CO_3$ would). Choice C makes the opposite error, suggesting $[CO_3^{2-}] = 2s$, as if each dissolved unit produced two carbonate ions. Choice D shows $[Ca^{2+}] = s^2$, which confuses molar solubility with the solubility product constant ($K_{sp}$) calculation. Always trace the stoichiometry from the balanced equation: count how many of each ion comes from one formula unit of the dissolving compound. For 1:1 stoichiometry like this, ion concentrations equal molar solubility. This relationship changes only when compounds have different ion ratios, like $Ca(OH)_2$ producing two hydroxide ions per formula unit.

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