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Standard XII

Chemistry

Solubility Product

The solubilit...

Question

The solubility of $S r (O H)_{2}$ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

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Solution

Solubility of $S r (O H)_{2} = 19.23 g / L$
Then, concentration of $S r (O H)_{2}$
$= \frac{19.23}{121.63} M = 0.1581 M S r (O H)_{2 (a q)} \to S r_{(a q)}^{2 +} + 2 (O H^{-})_{(a q)} ∴ [S r^{2 +}] = 0.1581 M [O H^{-}] = 2 \times 0.158 m = 0.3126 m$
Now,
$K_{w} = [O H^{-}] [H^{+}] \frac{10^{- 14}}{0.3126} = [H^{+}] \Rightarrow [H^{+}] = 3.2 \times 10^{- 14} ∴ p H = 13.495 (o r) 13.50$

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Solubility Product

Standard XII Chemistry

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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=19.23g/L Then, concentration of Sr(OH)2 =19.23121.63M=0.1581MSr(OH)2(aq)→Sr2+(aq)+2(OH−)(aq)∴[Sr2+]=0.1581M[OH−]=2×0.158m=0.3126m Now, Kw=[OH−][H+]10−140.3126=[H+]⇒[H+]=3.2×10−14∴pH=13.495(or)13.50

The solubility of $S r (O H)_{2}$ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.