Chemistry - Ionic Equilibrium Question with Solution | TestHub

Given:

Volume of $\mathrm{Ba}({\mathrm{NO}}_3{)}_2$ solution = $25.0$mL  Concentration of $\mathrm{Ba}({\mathrm{NO}}_3{)}_2$ solution = $0.050$M Volume of $\mathrm{NaF}$ solution = $25.0$mL  Concentration of $\mathrm{NaF}$ solution = $0.020$M

$\left[{\mathrm{Ba}}^{+2}\right]=\left(\frac{25\times 0.05}{50}\right)=0.025\mathrm{M}$

$\left[{\mathrm{F}}^{-}\right]=\left(\frac{25\times 0.02}{50}\right)=0.01\mathrm{M}$

$\left[{\mathrm{Ba}}^{+2}\right]$ $\left[{\mathrm{F}}^{-}\right]=25\times {10}^{-7}$

${\mathrm{K}}_{\mathrm{sp}}=0.5\times {10}^{-6}=5\times {10}^{-7}$

Ratio=$\frac{\left[{\mathrm{Ba}}^{+2}\right]\left[{\mathrm{F}}^{-}\right]}{K_{sp}}=\frac{25\times {10}^{-7}}{5\times {10}^{-7}}=5$