Question

$K_{sp}$ of $BaCr{O}_4$ is $2\times {10}^{-10}$ at room temperature. Calculate the mass of Barium chromate lost by washing with three litres of water. Will the loss in mass depend on the quantity of precipitate?

• A. 5.64 mg
• B. 10.625 mg
• C. 3.88 mg
• D. 15.6 mg

Answer 1

The correct option is B

10.625 mg

$BaCr{O}_4\left(s\right)$ $\rightleftharpoons$ $B{a}^{2+}+Cr{O}_4^{2-}$

$K_{SP}$ = $2\times {10}^{-10}$

$\therefore$ Solubility of $BaCr{O}_4$ = $\sqrt{2\times {10}^{-19}}$ = $1.41\times {10}^{-3}$

Concentration of barium chromate = $1.4\times {10}^{-3}$

M. Mass of $BaCr{O}_4$ = 253

$1.4\times {10}^{-7}$ moles of $BaCr{O}_4$ = $1.4\times {10}^{-3}\times 253g$

= $1.4\times {10}^{-3}\times 253\times 1000$ = $3.542mg$

for a single washing with one litre of water the mass of $BaCr{O}_4$ lost = $3.542mg$

mass of Barium Chromate lost by washing with three litres = $3.542\times 3$ = $10.626mg$

Loss of mass does not depend upon the quantity of the precipitate.