Question

How much $NaF$ should be added to $100$ mL of solution having $0.016M$ in $S{r}^{2+}$ ions to reduce its concentration to $2.5\times {10}^{-3}M$? ($K_{sp}Sr{F}_2$= $8\times {10}^{-10}$)

• A. $0.098$ g
• B. $0.168$ g
• C. $0.177$ g
• D. $0.118$ g

Answer 1

Answer: (D)

$0.118$ g

$\left[S{r}^{2+}\right]=2.5\times {10}^{-3}$ and $K_{sp}=8\times {10}^{-10}$

${Sr{F}_2}_{\left(s\right)}⟶S{r}^{2+}+2F^{-}$

$K_{sp}=\left[S{r}^{2+}\right][2F^{-}{]}^2$

$8\times {10}^{-10}=2.5\times {10}^{-3}\times \left[2F^{-}\right]2$

$\therefore \left[2F^{-}\right]=5.65\times {10}^{-4}$

$\therefore \left[F^{-}\right]=2.825\times {10}^{-4}$

This is the concentration in $1$ $L$ i.e., No. of moles in $1$ $L$.

$\therefore$ Concentration in $100$ $mL=\frac{2.825\times {10}^{-4}}{{10}^{-1}}=2.825\times {10}^{-3}$

Molecular weight of $NaF=42$

$\therefore$ Quantity of $NaF$ to be added $=42\times 2.825\times {10}^{-3}=0.118$ $g$