JEE Advanced · Chemistry · 8. Ionic Equilibrium
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 \(\boldsymbol{X} \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3}\). The value of \(\boldsymbol{X}\) is \(\qquad\) .
Use: Solubility product constant \(\left(K_{\mathrm{sp}}\right)\) of barium iodate \(=1.58 \times 10^{-9}\)
- A 3.95
- B 4.95
- C 5.95
- D 6.95
Answer & Solution
Correct Answer
(A) 3.95
Step-by-step Solution
Detailed explanation
\(\left[\mathrm{NaIO}_3\right]=\frac{6}{300}=2 \times 10^{-2} \mathrm{M}\)
\(\begin{aligned} \mathrm{Ba}\left(\mathrm{IO}_3\right)_2(\mathrm{~s}) \rightleftharpoons \mathrm{Ba}^{2+}+ & 2 \mathrm{IO}_3^{-} \\ \mathrm{s} \qquad & \left(2 \times 10^{-2}+2 \mathrm{~s}\right)\end{aligned}\)
\(\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{+2}\right]\left[\mathrm{IO}_3^{-}\right]^2 \\ & 1.58 \times 10^{-9}=\left[\mathrm{Ba}^{+2}\right] \times\left(2 \times 10^{-2}\right)^2 \\ & {\left[\mathrm{Ba}^{+2}\right]=\mathrm{s}=3.95 \times 10^{-6} \mathrm{M}} \\ & \mathrm{X}=3.95\end{aligned}\)
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