The $\{K_{sp}\}$ of $\{Zn(OH)_2\}$ is $3 \times 10^{-17}$. Determine its solubility in $0.01$ M $NaOH$.

## CORE INFORMATION\n✓ **Answer**: Option A\n## CONCEPT FRAMEWORK\n **Primary Concept**: Solubility product and common ion effect\n **Related Topics**: Equilibrium, solubility, ionic compounds\n **Prerequisites**: Understanding of equilibrium constants, solubility product ($K_{sp}$), and the common ion effect.\n## SOLUTION PATHWAY\n**Given Information**:\n $K_{sp} \text{ of } Zn(OH)_2 = 3 \times 10^{-17}$\n $[NaOH] = 0.01 M$\n $[OH^-] = 0.01 M$ (since NaOH is a strong base and completely dissociates)\n**Method & Steps**:\n 1. Write the equilibrium expression for the dissolution of $Zn(OH)_2$:\n $Zn(OH)_2(s) \rightleftharpoons Zn^{2+}(aq) + 2OH^-(aq)$\n $K_{sp} = [Zn^{2+}][OH^-]^2$\n 2. Let 's' be the molar solubility of $Zn(OH)_2$. Then, at equilibrium:\n $[Zn^{2+}] = s$\n $[OH^-] = 0.01 + 2s$ (This accounts for the hydroxide ions from NaOH and the dissolution of $Zn(OH)_2$)\n Since $K_{sp}$ is very small, we can assume that $2s << 0.01$, so $[OH^-] ≈ 0.01 M$.\n 3. Substitute the values into the $K_{sp}$ expression:\n $3 \times 10^{-17} = s(0.01)^2$\n $s = \frac{3 \times 10^{-17}}{(0.01)^2} = \frac{3 \times 10^{-17}}{10^{-4}} = 3 \times 10^{-13} M$\n Result: The solubility of $Zn(OH)_2$ in 0.01 M NaOH is $3 \times 10^{-13} M$\n Verification: The assumption that $2s << 0.01$ is valid since $2(3 \times 10^{-13}) << 0.01$.\n ## OPTION ANALYSIS\nA] ✓: $3 \times 10^{-13}$ is the correct solubility calculated using the $K_{sp}$ expression and the common ion effect.\nB] :x: $3 \times 10^{-17}$ is the value of $K_{sp}$, not the solubility.\nC] :x: This value is not supported by the calculation.\nD] :x: This value is not supported by the calculation.\n ## LEARNING AIDS\n **Common Mistakes**:\n 1.Ignoring the common ion effect: Failing to consider the hydroxide ions from NaOH affects the solubility calculation.\n How to avoid: Always account for the common ion when calculating solubility in the presence of a common ion.\n 2.Incorrectly simplifying the equilibrium expression: Incorrectly simplifying the [OH-] term can lead to an inaccurate solubility value.\n How to avoid: Check the validity of any assumptions made during simplification.\n **Quick Tips**:\n 1.Always write the balanced equilibrium reaction and the corresponding $K_{sp}$ expression before solving.\n 2.Remember that the common ion effect reduces the solubility of a sparingly soluble salt.

If the solubility of $\{Mg(OH)_2\}$ in pure water is $1.6 \times 10^{-4}$ M, what is its $\{K_{sp}\}$?

## CORE INFORMATION\n✓ **Answer**: Option A\n## CONCEPT FRAMEWORK\n **Primary Concept**: Solubility product constant ($K_{sp}$)\n **Related Topics**: Solubility, equilibrium constant, ionic compounds\n **Prerequisites**: Understanding of equilibrium expressions and molar solubility.\n## SOLUTION PATHWAY\n**Given Information**:\n Solubility of $Mg(OH)_2$ = $1.6 \times 10^{-4}$ M\n [Important conditions]: The solubility is in pure water, implying no common ion effect.\n**Method & Steps**:\n 1. Write the balanced dissolution equation for $Mg(OH)_2$:\n $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$\n 2. Write the expression for the solubility product constant ($K_{sp}$):\n $K_{sp} = [Mg^{2+}][OH^-]^2$\n 3. Determine the equilibrium concentrations of the ions:\n Since the solubility of $Mg(OH)_2$ is $1.6 \times 10^{-4}$ M, the equilibrium concentration of $Mg^{2+}$ is $1.6 \times 10^{-4}$ M, and the equilibrium concentration of $OH^-$ is $2 \times (1.6 \times 10^{-4})$ M = $3.2 \times 10^{-4}$ M.\n 4. Substitute the equilibrium concentrations into the $K_{sp}$ expression:\n $K_{sp} = (1.6 \times 10^{-4})(3.2 \times 10^{-4})^2$\n $K_{sp} = (1.6 \times 10^{-4})(10.24 \times 10^{-8})$\n $K_{sp} = 16.384 \times 10^{-12} \approx 1.64 \times 10^{-11}$\n 3. Final step:\n Result: $K_{sp} = 1.64 \times 10^{-11}$\n Verification: The units of $K_{sp}$ are consistent with the expression and the given solubility.\n ## OPTION ANALYSIS\nA] ✓: $1.64 \times 10^{-11}$ is the correct value calculated from the solubility.\nB] :x: $6.56 \times 10^{-12}$ is incorrect; it's likely a result of an error in the calculation of the hydroxide ion concentration or the $K_{sp}$ expression.\nC] :x: $2.56 \times 10^{-8}$ is incorrect; it might stem from an incorrect squaring of the hydroxide ion concentration.\nD] :x: $4.096 \times 10^{-16}$ is incorrect; this value might be obtained by incorrectly multiplying the concentrations instead of raising the hydroxide concentration to the power of 2.\n ## LEARNING AIDS\n **Common Mistakes**:\n 1.Incorrectly writing the $K_{sp}$ expression: Remember to raise the concentration of the hydroxide ion to the power of 2, as there are two hydroxide ions in the formula.\n 2.Forgetting to account for the stoichiometry: The concentration of hydroxide ions is twice the concentration of magnesium ions in this case.\n **Quick Tips**:\n 1.Always write the balanced dissolution equation before writing the $K_{sp}$ expression.\n 2.Double-check your calculations, especially when dealing with exponents.

The correct Option is "1.64e-11"

In a saturated solution of $\{Cd(OH)_2\}$ with a $\{K_{sp}\}$ of $2.5 \times 10^{-14}$, what is the concentration of $\{Cd^{2+}\}$ ions?

## CORE INFORMATION\n✓ **Answer**: Option A\n## CONCEPT FRAMEWORK\n **Primary Concept**: Solubility product constant ($K_{sp}$) and its application in calculating ion concentrations in saturated solutions.\n **Related Topics**: Equilibrium, solubility, ionic compounds.\n **Prerequisites**: Understanding of equilibrium constants, molarity, and stoichiometry.\n## SOLUTION PATHWAY\n**Given Information**:\n $K_{sp} \text{ of } Cd(OH)_2 = 2.5 \times 10^{-14}$\n The solution is saturated.\n**Method & Steps**:\n 1. Write the dissociation equation for $Cd(OH)_2$ in water:\n $Cd(OH)_2(s) \rightleftharpoons Cd^{2+}(aq) + 2OH^-(aq)$\n Reasoning: A saturated solution is at equilibrium, and the $K_{sp}$ expression represents this equilibrium.\n Formula/Rule: $K_{sp} = [Cd^{2+}][OH^-]^2$\n 2. Let $s$ be the molar solubility of $Cd(OH)_2$. Then, at equilibrium, $[Cd^{2+}] = s$ and $[OH^-] = 2s$. Substitute these into the $K_{sp}$ expression:\n Working: $2.5 \times 10^{-14} = (s)(2s)^2 = 4s^3$\n Key Point: The stoichiometry of the dissociation is crucial; for every mole of $Cd(OH)_2$ that dissolves, one mole of $Cd^{2+}$ and two moles of $OH^-$ are produced.\n 3. Solve for $s$:\n $s^3 = \frac{2.5 \times 10^{-14}}{4} = 6.25 \times 10^{-15}$\n $s = \sqrt[3]{6.25 \times 10^{-15}} \approx 1.84 \times 10^{-5}$\n Result: $[Cd^{2+}] = s = 1.84 \times 10^{-5} \text{ M}$\n Verification: Substitute the value of s back into the $K_{sp}$ expression to check: $(1.84 \times 10^{-5})(2(1.84 \times 10^{-5}))^2 \approx 2.5 \times 10^{-14}$\n ## OPTION ANALYSIS\nA] ✓: Correct. This is the calculated molar solubility of $Cd^{2+}$ ions.\nB] :x: Incorrect. This is the $K_{sp}$ value, not the concentration of $Cd^{2+}$ ions.\nC] :x: Incorrect. This value is not obtained from the correct calculation.\nD] :x: Incorrect. This value is not obtained from the correct calculation.\n ## LEARNING AIDS\n **Common Mistakes**:\n 1. Ignoring the stoichiometry of the dissociation: Forgetting to square the hydroxide ion concentration in the $K_{sp}$ expression leads to an incorrect answer.\n How to avoid: Always write the balanced dissociation equation and use the stoichiometric coefficients correctly in the $K_{sp}$ expression.\n 2. Incorrectly solving the cubic equation: Errors in solving for $s$ can arise. \n How to avoid: Carefully perform the mathematical operations and use a calculator if needed.\n **Quick Tips**:\n 1. Remember the general form of the $K_{sp}$ expression for a salt $A_xB_y$: $K_{sp} = [A]^x[B]^y$\n 2. Always check your answer by substituting it back into the $K_{sp}$ expression.

The correct Option is "1.84e-5"

In the following reactions, identify the conjugate bases:\n1. $H_2O \rightarrow H^+ + ?$ \n2. $HF \rightarrow H^+ + ?$

## CORE INFORMATION\n✓ **Answer**: B. $OH^-$ and $F^-$ \n## CONCEPT FRAMEWORK\n **Primary Concept**: Bronsted-Lowry Acid-Base Theory\n **Related Topics**: Conjugate acid-base pairs, proton transfer\n **Prerequisites**: Understanding of acids and bases, ability to identify proton transfer.\n## SOLUTION PATHWAY\n**Given Information**:\n Reactions: $H_2O \rightarrow H^+ + ?$ and $HF \rightarrow H^+ + ?$\n**Method & Steps**:\n 1. Identify the proton ($H^+$) donated by each acid.\n Reasoning: A Bronsted acid donates a proton.\n Formula/Rule: Acid + Base $\rightarrow$ Conjugate base + Conjugate acid\n 2. Determine the remaining species after proton removal to find the conjugate base.\n Working: For $H_2O$, removing $H^+$ leaves $OH^-$. For $HF$, removing $H^+$ leaves $F^-$.\n Key Point: The conjugate base has one less proton than the acid.\n 3. Match the conjugate bases to the given options.\n Result: $OH^-$ and $F^-$ are the conjugate bases of $H_2O$ and $HF$ respectively.\n Verification: Check if adding a proton to each conjugate base regenerates the original acid.\n ## OPTION ANALYSIS\nA :x: Incorrect. $H_3O^+$ is the conjugate acid of $H_2O$, not the conjugate base.\nB ✓: Correct. $OH^-$ is the conjugate base of $H_2O$, and $F^-$ is the conjugate base of $HF$.\nC :x: Incorrect. $H_3O^+$ is the conjugate acid of $H_2O$, and $H_2F^+$ is not a commonly known species.\nD :x: Incorrect. $O^{2-}$ requires removal of two protons from $H_2O$.\n ## LEARNING AIDS\n **Common Mistakes**:\n 1.Confusing conjugate acids and conjugate bases: Remember that the conjugate base has one less proton than the acid, and the conjugate acid has one more proton than the base.\n 2.Incorrectly removing protons: Carefully identify the proton that is donated by the acid.\n **Quick Tips**:\n 1.To find the conjugate base, simply remove a proton from the acid.\n 2.To find the conjugate acid, simply add a proton to the base.

The correct Option is "$OH^-$ and $F^-$"

If the solubility product ($K_{sp}$) of $Al(OH)_3$ is $27 \times 10^{-27}$ at 298 K, what is the pH of a saturated solution of $Al(OH)_3$?

## CORE INFORMATION\n✓ **Answer**: D. 7.73\n## CONCEPT FRAMEWORK\n **Primary Concept**: Solubility product and pH relationship\n **Related Topics**: Acid-base equilibrium, ionic product of water\n **Prerequisites**: Understanding of $K_{sp}$, pH, pOH, and equilibrium calculations.\n## SOLUTION PATHWAY\n**Given Information**:\n $K_{sp} = 27 \times 10^{-27}$\n $Al(OH)_3$ is a saturated solution.\n**Method & Steps**:\n 1. Write the dissociation equation and the expression for $K_{sp}$:\n $Al(OH)_3(s) \rightleftharpoons Al^{3+}(aq) + 3OH^-(aq)$\n $K_{sp} = [Al^{3+}][OH^-]^3$\n 2. Let $s$ be the molar solubility of $Al(OH)_3$. Then $[Al^{3+}] = s$ and $[OH^-] = 3s$.\n 3. Substitute into the $K_{sp}$ expression:\n $K_{sp} = s(3s)^3 = 27s^4$\n 4. Solve for $s$:\n $s^4 = \frac{K_{sp}}{27} = \frac{27 \times 10^{-27}}{27} = 10^{-27}$\n $s = (10^{-27})^{1/4} = 10^{-6.75} M$\n 5. Calculate $[OH^-]$:\n $[OH^-] = 3s = 3 \times 10^{-6.75} M$\n 6. Calculate pOH:\n $pOH = -log[OH^-] = -log(3 \times 10^{-6.75}) \approx 6.27$\n 7. Calculate pH:\n Reasoning: pH + pOH = 14\n pH = 14 - pOH = 14 - 6.27 = 7.73\n Result: pH = 7.73\n Verification: The pH is slightly basic, which is expected for a saturated solution of a metal hydroxide.\n ## OPTION ANALYSIS\nA :x: Incorrect calculation of pH.\nB :x: Incorrect calculation of pH.\nC :x: Incorrect calculation of pH.\nD ✓: Correct calculation of pH.\n ## LEARNING AIDS\n **Common Mistakes**:\n 1.Incorrectly substituting into the $K_{sp}$ expression.\n 2.Making errors in solving the equation for $s$.\n 3.Forgetting to consider the stoichiometry when calculating $[OH^-]$ from $s$.\n **Quick Tips**:\n 1.Always write the balanced dissociation equation and the $K_{sp}$ expression.\n 2.Remember that the molar solubility is related to the ion concentrations through the stoichiometry.

NEET Chemistry Chemical Equilibrium MCQs

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