This problem tests solubility equilibria (quantitative). The dissolution equation AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) shows a 1:1 stoichiometry, so if s moles of AgCl dissolve, we get s moles of Ag⁺ and s moles of Cl⁻. The Ksp expression is Ksp = [Ag⁺][Cl⁻] = (s)(s) = s². Solving: s² = 1.6×10⁻¹⁰, so s = √(1.6×10⁻¹⁰) = 1.26×10⁻⁵ M. Choice A (8.0×10⁻¹¹) incorrectly divides Ksp by 2 instead of taking the square root. For simple 1:1 salts like AgCl, the molar solubility equals the square root of Ksp.

This question tests your understanding of solubility equilibria (quantitative). The dissolution equation is BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq). To construct the ICE table, we set initial concentrations of Ba²⁺ and SO₄²⁻ to 0 M, the change as +x for both ions due to 1:1 stoichiometry, and equilibrium concentrations as x for both. The Ksp expression is [Ba²⁺][SO₄²⁻] = x · x = x² = $1.0×10^{-10}$, so we solve for the single variable x by taking the square root to get x = $1.0×10^{-5}$ M. We use only one variable because the equal stoichiometric coefficients link the ion concentrations directly. A tempting distractor is $1.0×10^{-20}$ M (choice E), which results from squaring the Ksp instead of taking the square root. For simple salts, Ksp problems reduce to one variable on AP MCQs, so always set up the expression based on stoichiometry and solve accordingly.

This problem tests solubility equilibria (quantitative). The dissolution equation $\text{AgCl}(s) \rightleftharpoons \text{Ag}^{+}(aq) + \text{Cl}^{-}(aq)$ shows a 1:1 stoichiometry, so if s moles of AgCl dissolve, we get s moles of Ag⁺ and s moles of Cl⁻. The Ksp expression is $K_{sp} = [\text{Ag}^{+}] [\text{Cl}^{-}] = (s)(s) = s^2$. Solving: $s^2 = 1.6\times10^{-10}$, so $s = \sqrt{1.6\times10^{-10}} = 1.26\times10^{-5} , \text{M}$. Choice A ($8.0\times10^{-11}$) incorrectly divides Ksp by 2 instead of taking the square root. For simple 1:1 salts like AgCl, the molar solubility equals the square root of Ksp.

This problem tests solubility equilibria (quantitative). The dissolution equation $\text{Fe(OH)}2(s) \rightleftharpoons \text{Fe}^{2+}(aq) + 2\text{OH}^{-}(aq)$ shows that dissolving s moles produces s moles of $\text{Fe}^{2+}$ and 2s moles of $\text{OH}^{-}$. The $K{sp}$ expression is $K_{sp} = [\text{Fe}^{2+}][\text{OH}^{-}]^2 = s \cdot(2s)^2 = 4s^3$. Solving: $4s^3 = 4.0 \times 10^{-14}$, so $s^3 = 1.0 \times 10^{-14}$, and $s = 1.0 \times 10^{-5} , \text{M}$, making $[\text{OH}^{-}] = 2s = 2.0 \times 10^{-5} , \text{M}$. Choice C ($4.0 \times 10^{-14}$) incorrectly uses the Ksp value as the concentration. For hydroxide salts, remember to multiply the molar solubility by 2 to get the OH⁻ concentration.

This problem tests solubility equilibria (quantitative). The dissolution equation CuCl(s) ⇌ Cu⁺(aq) + Cl⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Cu⁺ and Cl⁻. The Ksp expression is Ksp = [Cu⁺][Cl⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻⁶, so s = √(1.0×10⁻⁶) = 1.0×10⁻³ M, which equals [Cl⁻]. Choice A (1.0×10⁻⁶) incorrectly uses Ksp as the concentration. For any 1:1 salt, both ion concentrations equal the molar solubility at equilibrium.

This problem tests solubility equilibria (quantitative). The dissolution equation ZnS(s) ⇌ Zn²⁺(aq) + S²⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Zn²⁺ and S²⁻. The Ksp expression is Ksp = [Zn²⁺][S²⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻²⁴, so s = √(1.0×10⁻²⁴) = 1.0×10⁻¹² M. Choice B (1.0×10⁻²⁴) incorrectly uses the Ksp value as the molar solubility without taking the square root. For extremely insoluble salts like ZnS, the molar solubility is still found by taking the square root of Ksp.

This problem tests solubility equilibria (quantitative). The dissolution equation Fe(OH)₂(s) ⇌ Fe²⁺(aq) + 2OH⁻(aq) shows that dissolving s moles produces s moles of Fe²⁺ and 2s moles of OH⁻. The Ksp expression is Ksp = [Fe²⁺][OH⁻]² = (s)(2s)² = 4s³. Solving: 4s³ = 4.0×10⁻¹⁴, so s³ = 1.0×10⁻¹⁴, and s = 1.0×10⁻⁵ M, making [OH⁻] = 2s = 2.0×10⁻⁵ M. Choice C (4.0×10⁻¹⁴) incorrectly uses the Ksp value as the concentration. For hydroxide salts, remember to multiply the molar solubility by 2 to get the OH⁻ concentration.

This problem tests solubility equilibria (quantitative). The dissolution equation BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Ba²⁺ and SO₄²⁻. The Ksp expression is Ksp = [Ba²⁺][SO₄²⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻¹⁰, so s = √(1.0×10⁻¹⁰) = 1.0×10⁻⁵ M. Choice A (1.0×10⁻¹⁰) incorrectly uses Ksp as the concentration without solving. For 1:1 salts, always take the square root of Ksp to find the molar solubility.

This question tests your understanding of solubility equilibria (quantitative). The dissolution equation is AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). To construct the ICE table, we set initial concentrations of Ag⁺ and Cl⁻ to 0 M, the change as +x for both ions due to 1:1 stoichiometry, and equilibrium concentrations as x for Ag⁺ and x for Cl⁻. The Ksp expression is [Ag⁺][Cl⁻] = x · x = x² = $1.6×10^{-10}$, so we solve for the single variable x by taking the square root to get x ≈ $1.3×10^{-5}$ M. We use only one variable because the equal stoichiometric coefficients link the ion concentrations directly. A tempting distractor is $1.6×10^{-10}$ M (choice A), which results from mistakenly using the Ksp value directly without solving the equation. For simple salts, Ksp problems reduce to one variable on AP MCQs, so always set up the expression based on stoichiometry and solve accordingly.

This problem tests your understanding of solubility equilibria (quantitative). When CuBr dissolves, it produces equal moles of Cu⁺ and Br⁻, so if s = molar solubility, then [Cu⁺] = [Br⁻] = s at equilibrium. The Ksp expression is Ksp = [Cu⁺][Br⁻] = s × s = s². Substituting: 4.0 × 10⁻⁸ = s², so s = √(4.0 × 10⁻⁸) = 2.0 × 10⁻⁴ M. A common error is using the Ksp value directly or miscalculating the square root. For 1:1 salts on AP exams, the molar solubility equals the concentration of each ion.