# Question 4eab6

Mar 20, 2015

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

$\text{HQuCl}$ however will dissociate in aqueous solution to give ${\text{HQu}}^{+}$ and ${\text{Cl}}^{-}$.

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

$H Q {u}_{\left(a q\right)}^{+} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s Q {u}_{\left(a q\right)} + {H}_{3} {O}_{\left(a q\right)}^{+}$

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

$p {K}_{a} + p {K}_{b} = p {K}_{w}$, or

$p {K}_{a} + p {K}_{b} = 14 \implies p {K}_{a} = 14 - p {K}_{b}$

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

${K}_{a} = {10}^{- p K a}$

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