Ncert Exercises & Solutions

7.58 The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Solubility of $Sr {(OH)}_{2}$ = 19.23 g/L

Then, the concentration of $Sr {(OH)}_{2}$

$= \frac{19.23}{121.63} M$
$= 0.1581 M$
$S r {(O H)}_{2 (a q)} \to S {r^{2 +}}_{(a q)} + 2 {(O H^{-})}_{(a q)}$
$∴ [S r^{2 +}] = 0.1581 M$
$[O H^{-}] = 2 \times 0.1581 M = 0.3126 M$
$N o w,$
$K_{w} = [O H^{-}] [H^{+}]$
$\frac{10^{- 14}}{0.3126} = [H^{+}]$
$\Rightarrow [H^{+}] = 3.2 \times 10^{- 14}$
$∴ p H = 1.495; 13.50$

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