Question

4. In this experiment, you will also examine the solubility of Ca(OH)2(s) in a 0.200 M CaCl2 solution. a. Will the solubility of Ca(OH)2(s) in 0.200 M CaCl2 be equal to, smaller than, or larger than the solubility you calculated in question 3b? Incorrect equal to Correct: smaller than Incorrect larger than You are correct. Your receipt no. is 163-9343 Help: Receipt Previous Tries b. Using your calculated Ksp value from your answer to 3c, calculate the Ca(OH)2 solubility (grams per liter) in a 0.200 M CaCl2 solution. g/L Submission not graded. Use more digits. Tries 3/99 Previous Tries

Answer 1

Solubility of sparingly soluble salt as Ca(OH)2 decreases in presence of a common ion as Ca2+

Ca(OH)2(s) <==> Ca2+(aq) + 2OH-(aq)

Adding Ca2+ from CaCl2 would increase concentration of Ca2+ on right handside which pushed the equilibrium to the left hansdie and forms more insoluble Ca(OH)2(s).

Ksp of Ca(OH)2 = 5.5 x 10^-6 = [Ca2+][OH-]^2

with,

[Ca2+] = 0.200 M from CaCl2

we have,

[OH-] = sq.rt.(Ksp/[Ca2+])

          = sq.rt.(5.5 x 10^-6/0.200)

          = 5.24 x 10^-3 M

So, Ca(OH)2 solubility = 5.24 x 10^-3 mol/L x 74.093 g/mol = 0.388 g/L