CHEMISTRY, M2 EQ-Bank 7 MC

A solution is prepared by dissolving 2.00 g of calcium hydroxide \(\ce{(Ca(OH)2)}\) in enough water to produce a 500.0 mL solution. What is the concentration of hydroxide ions in the final solution?

0.108 mol L\(^{-1}\)
0.0540 mol L\(^{-1}\)
0.0270 mol L\(^{-1}\)
0.0162 mol L\(^{-1}\)

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\(A\)

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\(n\ce{(Ca(OH)2)} = \dfrac{2.00}{40.08 + 2(1.008) + 2(16.00)} = 0.0270\ \text{mol}\)

\(\ce{Ca(OH)2(s) -> Ca^{2+}(aq) + 2OH-(aq)}\)

Using molar ratios, \(n\ce{(OH^-)} = 0.0270 \times 2 = 0.0540\ \text{mol}\)

\(c\ce{(OH^-)} = \dfrac{n}{V} = \dfrac{0.0540}{0.5} = 0.108\ \text{mol L}^{-1}\)

\(\Rightarrow A\)