The solubility of barium phosphate of molar mass ' M ' g $\mathrm{mol}^{-1}$ in water is $x \mathrm{~g}$ per…

The solubility of barium phosphate of molar mass ' M ' g $\mathrm{mol}^{-1}$ in water is $x \mathrm{~g}$ per 100 mL at 298 K . Its solubility product is $1.08 \times\left(\frac{x}{M}\right)^a \times(10)^{\mathrm{b}}$. The values of $a$ and $b$ respectively are
  1. 7,5
  2. 5,7
  3. 5,5
  4. 7,7

Solution

$\mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2 \longrightarrow 3 \mathrm{Ba}^{2+}+2 \mathrm{PO}_4{ }^{3-}$
$\begin{aligned} & \text { Solubility of Barium }=\left(\frac{\frac{\mathrm{x}}{\mathrm{M}}}{10^{-1}}\right) \\ & =\left(\frac{\mathrm{x}}{\mathrm{M}} \times 10\right) \mathrm{mol} / \mathrm{L} \\ & \mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2 \longrightarrow 3 \mathrm{Ba}^{2+}+2 \mathrm{PO}_4{ }^{3-} \\ & \left(3 \frac{\mathrm{x}}{\mathrm{M}} \times 10\right)\left(2 \frac{\mathrm{x}}{\mathrm{M}} \times 10\right) \\ & \mathrm{K}_{\mathrm{sp}}=\left(\frac{3 \mathrm{x}}{\mathrm{M}} \times 10\right)^3 \times\left(\frac{2 \mathrm{x}}{\mathrm{M}} \times 10\right)^2\end{aligned}$
$\begin{aligned} & =27 \times\left(\frac{\mathrm{x}}{\mathrm{M}}\right)^3 \times 10^3 \times 4 \times\left(\frac{\mathrm{x}}{\mathrm{M}}\right)^2 \times 10^2 \\ & =108 \times\left(\frac{\mathrm{x}}{\mathrm{M}}\right)^5 \times 10^5 \\ & \mathrm{~K}_{\mathrm{sp}}=1.08 \times\left(\frac{\mathrm{x}}{\mathrm{M}}\right)^5 \times 10^7 \\ & \therefore \quad \mathrm{a}=5 \\ & \quad \mathrm{~b}=7\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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