Question

Solubility of calcium phosphate (molecular mass, M) in water is \(W_g\) per 100 mL at \(25 ^\circ C\). Its solubility product at \(25 ^\circ C.\) will be approximately.

(1) \(10^7 \left(\frac {W}{M}\right)^3\)

(2) \(10^7\left (\frac {W}{M}\right)^5\)

(3) \(10^3 \left(\frac {W}{M}\right)^5\)

(4) \(10^5 \left(\frac {W}{M}\right)^5\)

Answer 1

Correct option is (2) \(10^7\left (\frac {W}{M}\right)^5\)

\(S = \frac {W \times 10}{M}\)

\(Ca_3 (PO_4)_2 (s) \rightleftharpoons \underset {3s}{3}Ca^{2+} (aq.) + 2\underset {2s}{P}O^{3-}_4 (aq.)\)

\(S = \frac {W\times 1000}{M\times 100}=\frac {W\times 10}{M}\)

\(K_{sp}=(3s)^3 (2s)^2\)

= 108 \(s^5\)

\(=108 \times 10^5 \times \left(\frac {W}{M}\right)^5\)

\(=1.08 \times 10^7 \left(\frac {W}{M}\right)^5\)