Question

Solubility product of radium sulphate is $4 \times 10^{-11}$.
What will be the solubility of $Ra^{2+}$ in $0.10\,M\,Na_2SO_4$?

• A. $4\times10^{-10}\,M$
• B. $2\times10^{-5}\,M$
• C. $4\times10^{-5}\,M$
• D. $2\times10^{-10}\,M$

Answer 1

Answer: (A)

$RaSO_4 \rightleftharpoons Ra^{2+} +SO^{2-}4$
$K_{sp}=\left[Ra^{2+}\right]\left[SO^{2-}_{4}\right]$
Concentration of $SO^{2-}_{4}$ from $Na_{2}SO_{4}=0.10\,M$
$Ra^{2+}=\frac{4\times10^{-11}}{0.10}$
$=4\times10^{-10}\,M$