Question #d1234

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Question #d1234

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Answer 1 · 2015-02-04T01:31:22

Sulfuric acid is a strong diprotic acid that dissociates in aqueous solution in two steps, each with a different acid dissociation constant

$H_{2} S O_{4 (a q)} + H_{2} O_{(l)} ⇌ H S O_{4 (a q)}^{-} + H_{3}^{+} O_{(a q)}$ $H_2SO_(4(aq)) + H_2O_((l)) rightleftharpoons HSO_(4(aq))^(-) + H_3^(+)O_((aq))$ , $K_{a 1} = 2.4 \cdot 10^{6}$ $K_(a1) = 2.4 * 10^(6)$

and

$H S O_{4 (a q)}^{-} + H_{2} O_{(l)} ⇌ S O_{4 (a q)}^{2 -} + H_{3}^{+} O_{(a q)}$ $HSO_(4(aq))^(-) + H_2O_((l)) rightleftharpoons SO_(4(aq))^(2-) + H_3^(+)O_((aq))$ , $K_{a 2} = 1.0 \cdot 10^{- 2}$ $K_(a2) = 1.0 * 10^(-2)$

Since no mention of these constants is made, I assume that the second step is ignored and sulfuric acid's dissociation is expressed as

$H_{2} S O_{4 (a q)} ⇌ 2 H_{(a q)}^{+} + S O_{4 (a q)}^{2 -}$ $H_2SO_(4(aq)) rightleftharpoons 2H_((aq))^(+) + SO_(4(aq))^(2-)$

One mole of sulfuric acid will produce twice as much moles of protons in aqueous solution, which means that

$[H^{+}] = 2 \cdot [H_{2} S O_{4}]$ $[H^(+)] = 2 * [H_2SO_4]$

SInce pH is calculated as

$p H_{sol} = - log ([H^{+}])$ $pH_("sol") = -log([H^(+)])$ , you can easily determine that

$[H^{+}] = 10_{sol}^{- p H_{sol}} = 10^{- 3.5} = 0.000316 M$ $[H^(+)] = 10^(-pH_("sol")) = 10^(-3.5) = "0.000316 M"$

This means that molarity of the sulfuric acid is

$[H_{2} S O_{4}] = \frac{[H^{+}]}{2} = \frac{0.000136 M}{2} = 1.58 \cdot 10^{- 4} M$ $[H_2SO_4] = ([H^(+)])/2 = "0.000136 M"/2 = 1.58 * 10^(-4)"M"$

Answer 2 · 2015-02-04T01:38:00

The concentration = $1.58 \times 10^{- 4} m o l / l$ $1.58xx10^(-4)mol//l$

$H_{2} S O_{4} \to 2 H^{+} + S O_{4}^{2 -}$ $H_2SO_4rarr2H^++SO_4^(2-)$

$[p H] = 3.5$ $[pH]=3.5$

So

$- log [H^{+}] = 3.5$ $-log[H^+]=3.5$

$log [H^{+}] = - 3.5$ $log[H^+]=-3.5$

$10^{- 3.5} = [H^{+}]$ $10^(-3.5)=[H^+]$

$[H^{+}] = 3.16 \times 10^{- 4} m o l / l$ $[H^+]=3.16xx10^(-4)mol//l$

Since sulfuric acid is 2 molar with respect to $H^{+}$ $H^+$ the concentration of $H_{2} S O_{4} = \frac{3.16 \times 10^{- 4}}{2} = 1.58 \times 10^{- 4} m o l / l$ $H_2SO_4=(3.16xx10^(-4))/2=1.58xx10^(-4)mol//l$

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