Question

$K_{sp}$ for $Sr{F}_2=2.8\times {10}^{-9}$ at ${25}^{o}C$. How much $NaF$ should be added to $100mL$ of solution having $0.016M$ in ${Sr}^{2+}$ ions to reduce its concentration to $2.5\times {10}^{-3}M$?

• A. $0.123g$
• B. $0.128g$
• C. $0.118g$
• D. $0.114g$

Answer 1

The correct option is C $0.118g$

$\left[{Sr}^{2+}\right]=16\times {10}^{-3}M$

Left $\left[{Sr}^{2+}\right]=2.5\times {10}^{-3}M$

$\left[{Sr}^{2+}\right]$ precipitated $=(16-2.5)\times {10}^{-3}=13.5\times {10}^{-3}M$

$\therefore$ $\left[F^{-}\right]$ needed for this precipitation $=2\times 13.5\times {10}^{-3}M$

$=27.0\times {10}^{-3}M(\because {Sr}^{2+}+2F^{-}⟶Sr{F}_2)$

Also,
$\left[{Sr}^{2+}\right]{\left[F^{-}\right]}^2=K_{{sp}_{Sr{F}_2}}=2.8\times {10}^{-9}$

$\therefore$ ${\left[F^{-}\right]}^2=\frac{2.8\times {10}^{-9}}{2.5\times {10}^{-3}}$

$\left[F^{-}\right]=1.058\times {10}^{-3}M$ i.e., the concentration of $F^{-}$ which will also, appear in solution state.

Thus, $\left[F^{-}\right]$ needed$=[27.0+1.058]\times {10}^{-3}M$

$\therefore$ $NaF$ needed for $1$ litre $28.058\times {10}^{-3}\times 42g$

$\therefore$ $NaF$ needed for $100mL=\frac{28.058\times {10}^{-3}\times 42}{10}=0.1178g$