The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium n

Question

The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is $X 	imes 10^{-6} \mathrm{~mol} \mathrm{~dm}^{-3}$. The value of $X$ is $\_\_\_\_$ .\ Use: Solubility product constant $\left(K_{	ext {sp }}ight)$ of barium iodate $= 1.58 	imes 10^{-9}$\

Professor PC Thomas Classes · Answer

(3.95) : \ Conc. of $\left[\mathrm{IO}_{3}^{-}ight]=\frac{6}{300}=2 	imes 10^{-2} \mathrm{~mol}$\ $\mathrm{dm}^{-3}$\ $\mathrm{dm}^{-3}$ $$ \mathrm{Ba}^{2+}+2 \mathrm{IO}_{3}^{-} $$ $K_{s p}=\left[\mathrm{Ba}^{2+}ight]\left[\mathrm{IO}_{3}{ }^{-}ight]^{2}$\ $1.58 	imes 10^{-9}=\left[\mathrm{Ba}^{2+}ight]\left(2 	imes 10^{-2}ight)^{2}$\ $\left[\mathrm{Ba}^{2+}ight]=\frac{1.58 	imes 10^{-9}}{4 	imes 10^{-4}}=3.95 	imes 10^{-6}$\ $\mathrm{~mol} \mathrm{dm} \mathrm{dm}^{-3}$\ $\mathrm{Ba}\left(\mathrm{IO}_{3}ight)_{2} ightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{IO}_{3}^{-}$