The solubility of a dibasic acid $M ({O H)}_{2}$ having molecular mass $121$ is $19.23 g / L$ at $298 K$ . calculate the concentration of $M^{2 +}$ and ${O H}^{-}$ ions and the $p H$ of solution.

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Solution

Mass of $M (O H)_{2} = 121$
Solubility= $19.23 g / L$
$∴$ Molarity of $M (O H)_{2} = \frac{19.23}{121} = 0.158 M$
$M (O H)_{2} \to M^{2 +} + 2 O H^{-}$
$∴ [M^{2 +}] = 0.158 M$
$\Rightarrow [O H^{-}] = 2 \times 0.158 = 0.316 M$
$\Rightarrow [H^{+}] [O H^{-}] = K_{w} = 10^{- 14}$
$∴ [H^{+}] = \frac{10^{- 14}}{0.316} = 3.16 \times 10^{- 14}$
$∴ p^{H} = - log (3.16 \times 10^{- 14})$
$= 14 - log 3.16 = 13.5$
$∴ [M^{2 +}] = 0.158 M$
$\Rightarrow [O H^{-}] = 0.316 M$ & $p H = 13.5$

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The solubility of a dibasic acid M(OH)2 having molecular mass 121 is 19.23 g/L at 298 K. calculate the concentration of M2+ and OH− ions and the pH of solution.

Mass of M(OH)2=121Solubility= 19.23g/L∴ Molarity of M(OH)2=19.23121=0.158MM(OH)2→M2++2OH−∴[M2+]=0.158M⇒[OH−]=2×0.158=0.316M⇒[H+][OH−]=Kw=10−14∴[H+]=10−140.316=3.16×10−14∴pH=−log(3.16×10−14)=14−log3.16=13.5∴[M2+]=0.158M⇒[OH−]=0.316M & pH=13.5

The solubility of a dibasic acid $M ({O H)}_{2}$ having molecular mass $121$ is $19.23 g / L$ at $298 K$ . calculate the concentration of $M^{2 +}$ and ${O H}^{-}$ ions and the $p H$ of solution.