Answer to Question #177609 in General Chemistry for o-l=

Question

Calcium hydroxide is only slightly soluble in water. The K_sp value of Ca(OH)₂ is 1.3 x 10^-6.

(a) Write the equation for the dissociation of calcium hydroxide in water and the solubility product constant (K_sp) expression for calcium hydroxide. 

(b) Calculate the solubility of calcium hydroxide. 

(c) How would the solubility of calcium hydroxide vary (more soluble, less soluble, or the same) if the salt were to be dissolved in a 0.10 M solution of calcium chloride (CaCl₂) rather than pure water? Justify. 

(d) Would precipitation of calcium hydroxide be expected if a 0.25 L sample of 0.100 M NaOH is mixed with 0.50 L of 0.300 M CaCl₂? Justify with a calculation. 

Answer 1

a)

Ca(OH)_2(s) = Ca²⁺_(aq) + 2OH^-_(aq)

                                  ^{S                 2S,                 S = Molar Solubility}

K_sp = [Ca²⁺][OH^-]² = 4S³ = 1.3*10^-6;

b)

$S = \sqrt[3]{K_{sp}/4}=\sqrt[3]{(1.3*10^{-6})/4} =$ 6.88*10^-3M

c)

Ca(OH)_2(s) =  Ca²⁺_(aq) + 2OH^-_(aq)

                                  ^{S+0.1                 2S,                 S = Molar Solubility}

K_sp = [Ca²⁺][OH^-]² = (S+0.1)*4S² = 1.3*10^-6;

Let’s say S << 0.1M, so (S+0.1)*4S² $\approx$ 0.4S².

S = $\sqrt{K_{sp}/0.4}=\sqrt{1.3*10^{-6}/0.4}=1.80*10^{-3}M.$

Ca(OH)₂ is less soluble in a 0.1M CaCl₂ solution than in pure water.

d)

$[Ca^{2+}][OH^-]^2=\dfrac{0.3M*0.5L}{0.5L+0.25L}*\dfrac{0.1M*0.25L}{0.5L+0.25L}^2=2.22*10^{-4}M>K_{sp}.$

Precipitation of Ca(OH)₂ is expected.