11. magnesium hydroxide, mg(oh)₂, has a ( k_{sp} = 6.3 imes 10^{-10} ). the solubility of mg(oh)₂ will be the lowest in 1.0 l of which of the following?
a. 0.10 m hcl
b. 0.10 m naoh
c. 0.10 m mgcl₂
d. pure h₂o

12. for which of the following salts would the relationship between molar solubility, s, in mol/l, and the value of ( k_{sp} ) be represented by the equation ( k_{sp} = 4s^3 )?
a. pbco₃
b. mg₃(po₄)₂
c. ag₂so₄
d. mns

Question

11. magnesium hydroxide, mg(oh)₂, has a ( k_{sp} = 6.3 	imes 10^{-10} ). the solubility of mg(oh)₂ will be the lowest in 1.0 l of which of the following?
a. 0.10 m hcl
b. 0.10 m naoh
c. 0.10 m mgcl₂
d. pure h₂o

12. for which of the following salts would the relationship between molar solubility, s, in mol/l, and the value of ( k_{sp} ) be represented by the equation ( k_{sp} = 4s^3 )?
a. pbco₃
b. mg₃(po₄)₂
c. ag₂so₄
d. mns

Accepted Answer

### Question 11 # Explanation: ## Step1: Recall Le Chatelier's Principle for solubility equilibrium. The dissolution of $\ce{Mg(OH)2}$ is $\ce{Mg(OH)2(s) <=> Mg^{2+}(aq) + 2OH^{-}(aq)}$. The solubility is affected by the common ion effect or reaction with ions. ## Step2: Analyze each option: - **Option a (0.10 M HCl):** $\ce{H+}$ reacts with $\ce{OH-}$ (from $\ce{Mg(OH)2}$ dissolution), shifting equilibrium right, increasing solubility. - **Option b (0.10 M NaOH):** Provides $\ce{OH-}$ (common ion). Let solubility be $s$. Then $[\ce{Mg^{2+}}] = s$, $[\ce{OH-}] = 0.10 + 2s \approx 0.10$ (since $s$ is small). $K_{sp} = s(0.10)^2 = 6.3 imes10^{-10} \implies s = \frac{6.3 imes10^{-10}}{0.01} = 6.3 imes10^{-8}$. - **Option c (0.10 M $\ce{MgCl2}$):** Provides $\ce{Mg^{2+}}$ (common ion). Let solubility be $s$. Then $[\ce{Mg^{2+}}] = 0.10 + s \approx 0.10$, $[\ce{OH-}] = 2s$. $K_{sp} = 0.10 imes(2s)^2 = 6.3 imes10^{-10} \implies 0.10 imes4s^2 = 6.3 imes10^{-10} \implies s^2 = \frac{6.3 imes10^{-10}}{0.4} = 1.575 imes10^{-9} \implies s \approx 3.97 imes10^{-5}$. - **Option d (pure $\ce{H2O}$):** Let solubility be $s$. $[\ce{Mg^{2+}}] = s$, $[\ce{OH-}] = 2s$. $K_{sp} = s(2s)^2 = 4s^3 = 6.3 imes10^{-10} \implies s^3 = \frac{6.3 imes10^{-10}}{4} = 1.575 imes10^{-10} \implies s \approx \sqrt[3]{1.575 imes10^{-10}} \approx 5.4 imes10^{-4}$. Comparing $s$ values: Option b has $s = 6.3 imes10^{-8}$, Option c has $\approx 3.97 imes10^{-5}$, Option d has $\approx 5.4 imes10^{-4}$. So lowest solubility in 0.10 M NaOH? Wait, wait, no—wait, in Option b, $[\ce{OH-}]$ is 0.10 M (from NaOH) plus 2s. Wait, maybe I miscalculated. Wait, no: when we have a common ion, the common ion is in excess. Wait, let's re - calculate Option b: Dissolution: $\ce{Mg(OH)2 <=> Mg^{2+} + 2OH-}$. Let $s$ be solubility (mol/L). Then $[\ce{Mg^{2+}}] = s$, $[\ce{OH-}] = 0.10 + 2s$. But since $K_{sp}$ is small, $2s \ll 0.10$, so $[\ce{OH-}] \approx 0.10$. Then $K_{sp} = s imes(0.10)^2 = 6.3 imes10^{-10} \implies s = \frac{6.3 imes10^{-10}}{0.01} = 6.3 imes10^{-8}$. Option c: $\ce{MgCl2}$ provides $\ce{Mg^{2+}} = 0.10$ M. So $[\ce{Mg^{2+}}] = 0.10 + s \approx 0.10$, $[\ce{OH-}] = 2s$. Then $K_{sp} = 0.10 imes(2s)^2 = 0.10 imes4s^2 = 0.4s^2 = 6.3 imes10^{-10} \implies s^2 = \frac{6.3 imes10^{-10}}{0.4} = 1.575 imes10^{-9} \implies s = \sqrt{1.575 imes10^{-9}} \approx 3.97 imes10^{-5}$. Option d: Pure water, $K_{sp} = 4s^3 = 6.3 imes10^{-10} \implies s^3 = \frac{6.3 imes10^{-10}}{4} = 1.575 imes10^{-10} \implies s = \sqrt[3]{1.575 imes10^{-10}} \approx 5.4 imes10^{-4}$. Option a: HCl reacts with $\ce{OH-}$, so equilibrium shifts right, solubility is higher than in pure water. Now, comparing the solubilities: Option b ($6.3 imes10^{-8}$) < Option c ($3.97 imes10^{-5}$) < Option d ($5.4 imes10^{-4}$) < Option a (higher than d). Wait, but wait—maybe I made a mistake. Wait, NaOH provides more $\ce{OH-}$ than $\ce{MgCl2}$ provides $\ce{Mg^{2+}}$? Wait, the concentration of common ion: in NaOH, $[\ce{OH-}] = 0.10$; in $\ce{MgCl2}$, $[\ce{Mg^{2+}}] = 0.10$. The stoichiometry: for $\ce{OH-}$, the exponent is 2, for $\ce{Mg^{2+}}$, exponent is 1. So let's recast the $K_{sp}$ expression. For $\ce{Mg(OH)2}$, $K_{sp} = [\ce{Mg^{2+}}][\ce{OH-}]^2$. In Option b: $[\ce{OH-}] = 0.10$, so $[\ce{Mg^{2+}}] = \frac{K_{sp}}{[\ce{OH-}]^2} = \frac{6.3 imes10^{-10}}{(0.10)^2} = 6.3 imes10^{-8}$ (this is the solubility, since 1 mol of $\ce{Mg(OH)2}$ gives 1 mol of $\ce{Mg^{2+}}$). In Option c: $[\ce{Mg^{2+}}] = 0.10$, so $[\ce{OH-}] = \sqrt{\frac{K_{sp}}{[\ce{Mg^{2+}}]}} = \sqrt{\frac{6.3 imes10^{-10}}{0.10}} = \sqrt{6.3 imes10^{-9}} \approx 7.94 imes10^{-5}$. Then solubility $s = \frac{[\ce{OH-}]}{2} \approx 3.97 imes10^{-5}$ (since 1 mol of $\ce{Mg(OH)2}$ gives 2 mol of $\ce{OH-}$). Ah, here's the mistake earlier: solubility is related to the ion concentration by the stoichiometry. So in Option b, solubility $s = [\ce{Mg^{2+}}] = 6.3 imes10^{-8}$ (since 1 mol $\ce{Mg(OH)2}$ gives 1 mol $\ce{Mg^{2+}}$). In Option c, solubility $s = \frac{[\ce{OH-}]}{2} \approx 3.97 imes10^{-5}$ (since 2 mol $\ce{OH-}$ per 1 mol $\ce{Mg(OH)2}$). So indeed, Option b has lower solubility than Option c. Wait, but why? Because the common ion $\ce{OH-}$ has a higher concentration (0.10 M) and is squared in the $K_{sp}$ expression, so the effect is stronger. So the solubility is lowest in 0.10 M NaOH? Wait, no—wait, let's check the calculations again. Wait, $K_{sp} = 6.3 imes10^{-10}$. In pure water: $K_{sp} = s imes(2s)^2 = 4s^3 \implies s = \sqrt[3]{\frac{6.3 imes10^{-10}}{4}} \approx \sqrt[3]{1.575 imes10^{-10}} \approx 5.4 imes10^{-4}$ M. In 0.10 M $\ce{MgCl2}$: $[\ce{Mg^{2+}}] = 0.10$ M. So $K_{sp} = 0.10 imes(2s)^2 \implies 0.10 imes4s^2 = 6.3 imes10^{-10} \implies s^2 = \frac{6.3 imes10^{-10}}{0.4} \implies s = \sqrt{\frac{6.3 imes10^{-10}}{0.4}} \approx \sqrt{1.575 imes10^{-9}} \approx 3.97 imes10^{-5}$ M. In 0.10 M NaOH: $[\ce{OH-}] = 0.10$ M. So $K_{sp} = s imes(0.10)^2 \implies s = \frac{6.3 imes10^{-10}}{0.01} = 6.3 imes10^{-8}$ M. In 0.10 M HCl: $\ce{H+}$ reacts with $\ce{OH-}$, so the equilibrium shifts right, and solubility is higher than in pure water (since $\ce{OH-}$ is removed, so more $\ce{Mg(OH)2}$ dissolves). So comparing the solubilities: $6.3 imes10^{-8}$ (Option b) < $3.97 imes10^{-5}$ (Option c) < $5.4 imes10^{-4}$ (Option d) < Option a (higher than d). So the lowest solubility is in 0.10 M NaOH? Wait, but wait—maybe I messed up the common ion effect. Wait, the common ion with the higher concentration and higher exponent in the $K_{sp}$ expression will have a stronger effect. Since $\ce{OH-}$ has an exponent of 2, and its concentration is 0.10 M, while $\ce{Mg^{2+}}$ has exponent 1 and concentration 0.10 M, the $\ce{OH-}$ (from NaOH) will suppress solubility more. So Option b (0.10 M NaOH) has the lowest solubility. Wait, but let's check again. Wait, another way: the solubility of $\ce{Mg(OH)2}$ in a solution with $[\ce{OH-}] = c$ is $s = \frac{K_{sp}}{c^2}$. In a solution with $[\ce{Mg^{2+}}] = c$, solubility is $s = \sqrt{\frac{K_{sp}}{c}}$. So for $c = 0.10$ M: - In NaOH (c = 0.10 M for $\ce{OH-}$): $s = \frac{6.3 imes10^{-10}}{(0.10)^2} = 6.3 imes10^{-8}$ M. - In $\ce{MgCl2}$ (c = 0.10 M for $\ce{Mg^{2+}}$): $s = \sqrt{\frac{6.3 imes10^{-10}}{0.10}} \approx 7.94 imes10^{-5}$ M (but since each $\ce{Mg(OH)2}$ gives 2 $\ce{OH-}$, the solubility is half of that? Wait, no—wait, solubility $s$ is the moles of $\ce{Mg(OH)2}$ dissolved per liter. So $[\ce{Mg^{2+}}] = s$, $[\ce{OH-}] = 2s$. So in $\ce{MgCl2}$ solution, $[\ce{Mg^{2+}}] = s + 0.10 \approx 0.10$, so $s \approx \frac{K_{sp}}{(2s)^2 imes 0.10}$? No, earlier approach was correct: $K_{sp} = [\ce{Mg^{2+}}][\ce{OH-}]^2$. If $[\ce{Mg^{2+}}] = 0.10$, then $[\ce{OH-}] = \sqrt{\frac{K_{sp}}{0.10}}$, and since $[\ce{OH-}] = 2s$, then $s = \frac{1}{2}\sqrt{\frac{K_{sp}}{0.10}} \approx \frac{1}{2} imes7.94 imes10^{-5} \approx 3.97 imes10^{-5}$ M. Which is still higher than $6.3 imes10^{-8}$ M (from NaOH). So yes, Option b has lower solubility. # Answer: b. 0.10 M NaOH ### Question 12 # Explanation: ## Step1: Recall the relationship between molar solubility ($s$) and $K_{sp}$ for a salt with formula $\ce{A_xB_y}$. The dissolution is $\ce{A_xB_y(s) <=> xA^{y+}(aq) + yB^{x-}(aq)}$. Then $K_{sp} = [\ce{A^{y+}}]^x [\ce{B^{x-}}]^y = (xs)^x (ys)^y = x^x y^y s^{x + y}$. ## Step2: Analyze each option: - **Option a ($\ce{PbCO3}$):** Formula $\ce{AB}$ (x = 1, y = 1). Dissolution: $\ce{PbCO3 <=> Pb^{2+} + CO3^{2-}}$. $K_{sp} = (s)(s) = s^2$. - **Option b ($\ce{Mg3(PO4)2}$):** Formula $\ce{A3B2}$ (x = 3, y = 2). Dissolution: $\ce{Mg3(PO4)2 <=> 3Mg^{2+} + 2PO4^{3-}}$. $[\ce{Mg^{2+}}] = 3s$, $[\ce{PO4^{3-}}] = 2s$. $K_{sp} = (3s)^3 (2s)^2 = 27s^3 imes 4s^2 = 108s^5$. - **Option c ($\ce{Ag2SO4}$):** Formula $\ce{A2B}$ (x = 2, y = 1). Dissolution: $\ce{Ag2SO4 <=> 2Ag+ + SO4^{2-}}$. $[\ce{Ag+}] = 2s$, $[\ce{SO4^{2-}}] = s$. $K_{sp} = (2s)^2 (s) = 4s^2 imes s = 4s^3$. - **Option d ($\ce{MnS}$):** Formula $\ce{AB}$ (x = 1, y = 1). Dissolution: $\ce{MnS <=> Mn^{2+} + S^{2-}}$. $K_{sp} = (s)(s) = s^2$. # Answer: c. $\ce{Ag2SO4}$

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QUESTION IMAGE

Question

Explanation:

Question 11

Step1: Recall Le Chatelier's Principle for solubility equilibrium. The dissolution of $\ce{Mg(OH)2}$ is $\ce{Mg(OH)2(s) <=> Mg^{2+}(aq) + 2OH^{-}(aq)}$. The solubility is affected by the common ion effect or reaction with ions.

Step2: Analyze each option:

Step1: Recall the relationship between molar solubility ($s$) and $K_{sp}$ for a salt with formula $\ce{A_xB_y}$. The dissolution is $\ce{A_xB_y(s) <=> xA^{y+}(aq) + yB^{x-}(aq)}$. Then $K_{sp} = [\ce{A^{y+}}]^x [\ce{B^{x-}}]^y = (xs)^x (ys)^y = x^x y^y s^{x + y}$.

Step2: Analyze each option:

Answer:

Question 12