Question

How many grams of ba(io3)2 can be dissolved in 250. Ml of distilled water at 25 oc ? Note: ksp of ba(io3)2 is 1.57 × 10-9 and molar mass of ba(io3)2 = 487.1323 g/mol?

Answer 1

$0.142 \; \text{g} \; \text{Ba}(\text{IO}_3)_2$

Explanation

Solid barium iodate and its ionic components are in a dynamic equilibrium when dissolved in an aqueous solution. Each mole of barium iodate $\text{Ba}(\text{IO}_3)_2$ dissolves to produce one mole of barium $\text{Ba}^{2+}$ions and two moles of iodate $\text{IO_3}^{-}$ ions.

$\text{Ba}(\text{IO}_3)_2 \; (aq) \leftrightharpoons \text{Ba}^{2+} \; (aq) + 2 \; \text{IO}_3^{-} \; (aq)$

Thus

$K_{\text{sp}} &= & [\text{Ba}^{2+}] \cdot [\text{IO}_3^{-}]^{2}$.

Given that

$K_{\text{sp}}$ of $1.57 \times 10^{-9}$ from the question, and

$[\text{Ba}(\text{IO}_3)_2, \; \text{dissolved}] = [\text{Ba}^{2+}] = 2 \; [\text{IO}_3^{-}]$ as seen in the dissociation equilibrium,

$\begin{array}{lll}[\text{Ba}(\text{IO}_3)_2, \; \text{dissolved}]^{3} &=& [\text{Ba}^{2+}] \cdot [\text{IO}_3^{-}]^{2} \\ & = & K_{sp}\\&=&1.57\times 10^{-9} \end{array}$

$[\text{Ba}(\text{IO}_3)_2, \; \text{dissolved}] = \sqrt[3]{K_{\text{sp}}} = 1.16 \times 10^{-3} \; \text{mol}\cdot \text{dm}^{-3}$

$250 \; \text{ml} = 0.250 \text{dm}^{-3}$ of distilled water can thus dissolve up to

$\begin{array}{lll} n &=& c \cdot V\\ &= & 1.16 \times 10^{-3} \; \text{mol} \; \text{Ba}(\text{IO}_3)_2\cdot \text{dm}^{-3} \times 0.250 \; \text{dm}^{-3} \\ & = & 2.91 \times 10^{-4} \; \text{mol} \; \text{Ba}(\text{IO}_3)_2\; \text{,}\end{array}$

which corresponds to

$\begin{array}{lll} m &= & n \cdot M\\ &= & 2.91 \times 10^{-4} \; \text{mol} \; \text{Ba}(\text{IO}_3)_2 \times 487.1323 \; \text{g} \cdot \text{mol}^{-1} \\ & = &0.142 \; \text{g} \; \text{Ba}(\text{IO}_3)_2 \end{array}$