Question

uestion 2 Calculate the chromium ion concentration in a saturated solution of chromium(IV) carbonate (Ksp = 8.1 - 10-14)

Answer 1

The dissociation of sparingly soluble salt Cr2(CO3)4 can be shown as

Cr2(CO3)4 $\rightleftharpoons$ 2Cr^4+ + 4CO3^2-

i.e one mole of Cr2(CO3)4 in the solution gives 2 moles of Cr^4+ and 4 moles of CO3^2- ions.

Let the solubility of Cr2(CO3)4 be S mole per litre.

Hence in the solution, [Cr^4+] = 2S molL^-1 , [CO3^2- ] = 4S molL^-1

Now applying law of solubility product,

Ksp = [Cr^4+]^2 [CO3^2- ]^4 = (2S)^2 (4S)^4 = 1024S^6 = 8.1 * 10^-14    (Given)

$\Rightarrow$ S = (7.91 * 10^ -18)^ 1/6

$\Rightarrow$ S = 1.4 * 10^-3 molL^-1

Now concentration of [Cr^4+] ions = 2S = 2*1.4 * 10^-3 molL^-1 = 2.81*10^-3 mol ion L^-1