Question

An unknown quantity of Ca(OH)2 is added to the pure water. At equilibrium, the final pH was found to be 9.2.

MM of Ca(OH)2=74g/mol

Calcium hydroxide: Ca(OH)2 -> Ca2+ + 2OH- Ksp: 2 x 10^-14  M³

Water dissociation: H2O -> H+ + OH Kw: 1.0 x 10^-14 M²

Is there any solid remaining at equilibrium? Explain quantitatively.

Answer 1

As per the solublity product calculation according to the equilibrium

Ca(OH)₂ leading to the formation of Ca²⁺ and 2 moles of OH^- the solubility product expression is S*2S*2S= 4 S³ which is equal to 2*10^-14 as mentioned. So, S= (cube root of 5) * 10^-5 = 1.709 *10^-5mol/L (Note: The concentration of OH^- supplied by water is negligible as compared to that by Calcium Hydroxide)

Now , pH of the solution being 9.2, the pOH of the soution is 14-9.2=4.8. Calculating the antilogarithm of this value gives us the concentration of OH^- to be 1.58*10^-5 mol/L.

We see that there is some amount of unreacted OH^- present in the solution which comes out to be (1.709-1.580) *10^-5 mol/L= 0.129* 10-5 mol/L.. So there is some amount of solid remaining at equilibrium.