Question

The solubility product of $BaCrO _{4}$ is $2.4 \times 10^{-10} M ^{2}$. The maximum concentration of $Ba \left( NO _{3}\right)_{2}$ possible without precipitation in a $6 \times 10^{-4} MK _{2} CrO _{4}$ solution is

• A. $4 \times 10^{-7} M$
• B. $1.2 \times 10^{10} M$
• C. $6 \times 10^{-4} M$
• D. $3 \times 10^{-4} M$

Answer 1

Answer: (A)

Precipitation occurs, if $Q _{ c }> K _{ sp }$. The maximum concentration of $Ba \left( NO _{3}\right)_{3}$ or $Ba 2+$ without precipitation is obtained if

$Q _{ c }= K _{ sp }=\left[ Ba ^{2+}\right]\left[ CrO _{4}^{2-}\right]$

$2.4 \times 10^{-10} M ^{2}=\left[ Ba ^{2+}\right]\left(6 \times 10^{-4} M \right)$

or $\left[ Ba ^{2+}\right]=\frac{2.4 \times 10^{-10} M ^{2}}{6 \times 10^{-4} M }$

$=0.4 \times 10^{-6} M$

$=4 \times 10^{-7} M$