Question

Using stock bottles of CaCl₂ and KOH solutions, a student prepared a 1.00L solution mixture which contained, before any reaction occurred, [Ca²⁺] = 0.075M and [OH^–] = 0.085M. After mixing the solution, a precipitate of Ca(OH)₂ (s) was formed (reaction occurred in the right-to-left direction). The student then added a further 1.00L of water and stirred; the precipitate disappeared— all the solid had redissolved.

Which one of the following is a possible value for the equilibrium constant, K_c, of this reaction?

Ca(OH)₂ (s) <---> Ca²⁺ (aq) + 2OH^– (aq), Kc = ?

a) 8.7 x 10^-4

b) 1.1 x 10^-2

c) 2.7 x 10^-5

d) 2.1 x 10^-4

e) The experimental results described are impossible

*The answer is D but I don't know how to get there. It says it's a reaction quotient problem and I calculated Q = 5.42 x 10^-4 so I know Kc would have to be smaller than that. Am I supposed to use an ICE table? How does it change when the precipitate dissolves? Do the concentrations of Ca and OH change after the extra water is added? I am very confused.

Answer 1