25 . 0mL of 0 . 050MBa ( NO _ 3 ) _ 2 is mixed with 25 . 0mL of 0 ....

Given:

Volume of $Ba ( {NO}_{3}{)}_{2}$ solution = $25 . 0$mL Concentration of $Ba ( {NO}_{3}{)}_{2}$ solution = $0 . 050$M Volume of $NaF$ solution = $25 . 0$mL Concentration of $NaF$ solution = $0 . 020$M

\left[{Ba}^{+ 2}\right] = \left(\dfrac{25 \times 0 . 05}{50}\right) = 0 . 025M$</p><p>\left[{F}^{-}\right] = \left(\dfrac{25 \times 0 . 02}{50}\right) = 0 . 01M$

\left[{Ba}^{+ 2}\right] \left[{F}^{-}\right] = 25 \times {10}^{- 7}$</p><p>${K}_{sp} = 0 . 5 \times {10}^{- 6} = 5 \times {10}^{- 7}$</p><p>Ratio=$\dfrac{\left[{Ba}^{+ 2}\right]\left[{F}^{-}\right]}{{K}_{sp}} = \dfrac{25 \times {10}^{- 7}}{5 \times {10}^{- 7}} = 5$