The Solubility product constant (Ksp) of Ca (OH)2 at 25°C is 4.42x10-5. A 500 ml of Saturated Solution of Ca(OH)2 is mixed with an equal Volume of 0.4m NaOH. What mass of Ca (OH)2, is Precipitated out?

Question Collected from Telegram Group.

The Solubility product constant (K_sp) of Ca (OH)₂ at 25°C is 4.42x10^-5. A 500 ml of Saturated Solution of Ca(OH)₂ is mixed with an equal Volume of 0.4m NaOH. What mass of Ca (OH)₂, is Precipitated out?

Solution:

Let S M be the solubility of Ca(OH)₂​ in pure water.

Ca(OH)2	⇌	Ca+2	+ 2OH−
S		S	2S

K_sp​=[Ca²⁺][OH⁻]²

Substitute values in the above expression.

4.42×10⁻⁵=S(2S)² or S=0.0223M

500 ml of solution contains 0.01115 moles of calcium ions.

The molarity of NaOH solution is 0.4 M. When it is mixed with equal volume of calcium hydroxide solution, its molarity is reduced to half, i.e., 0.2M.

[OH⁻]=0.2M

[Ca+²]= $\frac{K_{sp}}{[OH^-]^2}=\frac{4.42\times 10^{-5}}{{0.4}^2}=0.001105\text{ M}$

The number of moles of calcium ion precipitated =0.01115−0.001105=0.010045.
The number of millimoles of calcium hydroxide precipitated =0.010045×74×1000=743 mg.