Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 60.00 mL sample titrated with HCl requires 11.13 mL of 9.83×10−2 M HCl to reach the endpoint. calculate ksp for Ca(OH)2?
To calculate the (K_{sp}) for (Ca(OH)_2), we follow these steps:

Determine the amount of (HCl) used: (11.13 , \text{mL} \times 9.83 \times 10^{2} , \text{mol/L} = 1.094 \times 10^{3} , \text{mol} , HCl)

Write the neutralization reaction: (Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O). This shows that 2 moles of (HCl) react with 1 mole of (Ca(OH)_2).

Calculate moles of (Ca(OH)_2) in the sampled solution: Since it takes 2 moles of (HCl) to neutralize 1 mole of (Ca(OH)_2), the moles of (Ca(OH)_2) is half the moles of (HCl), which is (\frac{1.094 \times 10^{3}}{2} = 5.47 \times 10^{4} , \text{mol}).

Calculate the concentration of (Ca(OH)_2) in the solution: The volume of the sample is 60.00 mL or 0.06000 L. So, the concentration of (Ca(OH)_2) is (\frac{5.47 \times 10^{4} , \text{mol}}{0.06000 , \text{L}} = 9.12 \times 10^{3} , \text{M}).

Write the dissociation equation and expression for (K_{sp}): (Ca(OH)2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^{}(aq)). The (K{sp}) expression is (K_{sp} = [Ca^{2+}][OH^]^2).

Calculate the ion concentrations in the saturated solution: Since the concentration of (Ca(OH)_2) in solution is (9.12 \times 10^{3} , \text{M}), and each formula unit of (Ca(OH)_2) produces one (Ca^{2+}) ion and two (OH^{}) ions, the concentrations are:
 ([Ca^{2+}] = 9.12 \times 10^{3} , \text{M})
 ([OH^{}] = 2 \times 9.12 \times 10^{3} , \text{M} = 1.824 \times 10^{2} , \text{M})

Calculate (K_{sp}): Using the concentrations found, [ K_{sp} = (9.12 \times 10^{3})(1.824 \times 10^{2})^2 = 9.12 \times 10^{3} \times 3.327 \times 10^{4} = 3.03 \times 10^{6} ]
Therefore, the (K_{sp}) for (Ca(OH)_2) is (3.03 \times 10^{6}).
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