Question #4eab6

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Question #4eab6

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Answer 1 · 2015-03-20T11:17:43

No, $HQuCl$ $"HQuCl"$ is not the same as ${HQu}^{+}$ $"HQu"^(+)$ , since the former is a salt and the latter is the conjugate acid of your weak base, $Qu$ $"Qu"$ .

$HQuCl$ $"HQuCl"$ however will dissociate in aqueous solution to give ${HQu}^{+}$ $"HQu"^(+)$ and ${Cl}^{-}$ $"Cl"^(-)$ .

$H Q u C l_{(a q)} ⇌ H Q u_{(a q)}^{+} + C l_{(a q)}^{-}$ $HQuCl_((aq)) rightleftharpoons HQu_((aq))^(+) + Cl_((aq))^(""-)$

The reaction of interest will be the dissociation of the conjugate acid to form your original base and the hydronium ion

$H Q u_{(a q)}^{+} + H_{2} O_{(l)} ⇌ Q u_{(a q)} + H_{3} O_{(a q)}^{+}$ $HQu_((aq))^(+) + H_2O_((l)) rightleftharpoons Qu_((aq)) + H_3O_((aq))^(+)$

This is the reaction you'll use to calculate the pH of the solution. Do not forget to use $p K_{a}$ $pK_a$ , not $p K_{b}$ $pK_b$ , for this equilibrium. Remember that

$p K_{a} + p K_{b} = p K_{w}$ $pK_a + pK_b = pK_w$ , or

$p K_{a} + p K_{b} = 14 \Rightarrow p K_{a} = 14 - p K_{b}$ $pK_a + pK_b = 14 => pK_a = 14 - pK_b$

Use this value to determine the value of the acid dissociation constant, $K_{a}$ $K_a$ , by

$K_{a} = 10^{- p K a}$ $K_a = 10^(-pKa)$

Now you've got all you need to calculate the pH of the solution.

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