Equilibrium answer all components Justify with a calculation.

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)

Ksp = [Ca2+][OH-]²

(b) 1.3x10^-6 = (x)(2x)² = 4x³

x3 = 3.2510^-7

x 6.9x10^-4 M = solubility of Ca(OH)₂

(c) The solubility would be less than in pure water due to the common ion effect. According to LeChatelier, the common ion (Ca²⁺) would push the reaction to the left forming more insoluble Ca(OH)₂.

(d) 2NaOH(aq) + CaCl_2(aq)=> Ca(OH)_2(s)+ 2NaCl(aq)

moles NaOH = 0.25 L x 0.1 mol/L = 0.025 mol NaOH = moles OH-

moles CaCl2 = 0.50 L x 0.3 mol/L = 0.15 mol CaCl2

Final volume = 75 ml = 0.075 L

Final [Ca2+] = 0.15 mol/0.075 L = 2 M

Final [OH-] = 0.025 mol / 0.075 L = 0.333 M

Q = [Ca²⁺][OH-]² = (2)(0.333)² = 0.22 M

Q > Ksp so a precipitate will form