JEE Advanced 2025 — Paper 2 — Question 39

$\quad \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaIO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{~s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})$

$2 \mathrm{mmol} \quad 10 \mathrm{mmol}$

LR ${llll} - 6 mmol - 4 mmol$

$\left[\mathrm{NaIO}_{3}\right]=\frac{6}{300}=2 \times 10^{-2} \mathrm{M}$

$\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{IO}_{3}^{-}$

$$\text { s } \quad\left(2 \times 10^{-2}+2 \mathrm{~s}\right)$$

$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{+2}\right]\left[\mathrm{IO}_{3}\right]^{2}$

$1.58 \times 10^{-9}=\left[\mathrm{Ba}^{+2}\right] \times\left(2 \times 10^{-2}\right)^{2}$

$\left[\mathrm{Ba}^{+2}\right]=\mathrm{s}=3.95 \times 10^{-6} \mathrm{M}$

$\mathrm{X}=3.95$