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 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 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 problem tests solubility equilibria (quantitative). For BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq), both ions form in a 1:1 ratio, so [Ba²⁺] = [SO₄²⁻] = x at equilibrium. Substituting into Ksp = [Ba²⁺][SO₄²⁻] = x² = 1.0×10⁻¹⁰. Solving gives x = √(1.0×10⁻¹⁰) = 1.0×10⁻⁵ M. A common mistake is reporting the Ksp value (1.0×10⁻¹⁰ M) as the concentration. For simple 1:1 salts, molar solubility equals the square root of Ksp.

This problem tests your understanding of solubility equilibria (quantitative). When $\text{BaSO}_4$ dissolves, it produces equal moles of $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$, so if $s$ = molar solubility, then $[\text{Ba}^{2+}] = [\text{SO}4^{2-}] = s$ at equilibrium. The Ksp expression is $K{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = s \times s = s^2$. Substituting: $1.0 \times 10^{-10} = s^2$, so $s = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} , \text{M}$. A common error is confusing this 1:1 salt with more complex stoichiometries or using Ksp directly as the concentration. For simple 1:1 salts, the calculation reduces to taking the square root of Ksp.

This problem tests solubility equilibria (quantitative). For ZnS(s) ⇌ Zn²⁺(aq) + S²⁻(aq), both ions have 1:1 stoichiometry, so if 's' mol/L dissolves, then [Zn²⁺] = [S²⁻] = s at equilibrium. The Ksp expression is Ksp = [Zn²⁺][S²⁻] = s × s = s², so 1.0×10⁻²⁴ = s², giving s = 1.0×10⁻¹² mol/L. This equals the molar solubility since one formula unit produces one of each ion. A tempting error is to take the fourth root instead of square root (if confused with more complex salts). For 1:1 salts, the calculation is straightforward: molar solubility = √Ksp.

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). For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), we set up an ICE table where initially both ion concentrations are 0, and at equilibrium both equal 's' (the molar solubility) due to the 1:1 stoichiometry. The Ksp expression becomes Ksp = [Ag⁺][Cl⁻] = s × s = s², so 1.6×10⁻¹⁰ = s², giving s = 1.3×10⁻⁵ mol/L. A common error is using Ksp directly as the concentration (choice A), forgetting that Ksp equals the product of ion concentrations, not the individual concentrations. For 1:1 salts like AgCl, remember that molar solubility equals the square root of Ksp.

This problem tests solubility equilibria (quantitative). For CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq), the 1:1 stoichiometry means if 's' mol/L dissolves, then [Ca²⁺] = [CO₃²⁻] = s at equilibrium. The Ksp expression is Ksp = [Ca²⁺][CO₃²⁻] = s × s = s², so 9.0×10⁻⁹ = s², giving s = 3.0×10⁻⁵ mol/L. Since the question asks for [CO₃²⁻], which equals s, the answer is 3.0×10⁻⁵ mol/L. A common mistake is taking the wrong root or confusing the carbonate concentration with total dissolved amount. For 1:1 salts, the concentration of each ion at equilibrium equals the molar solubility.

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.