# Question #859e4

##### 1 Answer
Jun 4, 2015

The concentration of the hydronium ions will be largely determined by the presence of the hydrochloric acid, which is a strong acid that completely dissociates in aqueous solution.

By comparison, $H A$, which is a weak acid, will not ionize completely in aqueous solution. $H A$ will react with water according to the following equilibrium

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

The smaller the acid dissociation constant is, the more the equilibrium will lie to the left, meaning that fewer molecules of $H A$ will lose their proton in solution.

Moreover, the presence of the hydrochloric acid will push this equilibrium even further to the left, because of the relatively large concentration of hydronium ions that results from its dissociation - this is known as Le Chatelier's Principle.

So, when hydrochloric acid dissociates, it forms hydronium ions and chloride ions

$H C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} \to {H}_{3} {O}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

Since you have a $1 : 1$ mole ratio between hydrochloric acid and the hydronium ion, every mole of the former will produce 1 mole of the latter. Thus,

$\left[{H}_{3} {O}^{+}\right] = \left[H C l\right] = \text{0.0100 M}$

Now examine what happens with the weak acid.

$\text{ } H {A}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + {A}_{\left(a q\right)}^{-}$
I.......0.10.................................0.0100.........0
C......(-x)......................................(+x)..........(+x)
E......0.10-x............................0.0100+x........x

The acid dissociation constant will be equal to

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[{A}^{-}\right]}{\left[H A\right]} = \frac{\left(0.0100 + x\right) \cdot x}{0.10 - x}$

Since ${K}_{a}$ has such a small value, you can approximate (0.0100 + x) and (0.10 - x) with 0.0100 and 0.10, respectively. This will get you

${K}_{a} = \frac{0.0100}{0.10} \cdot x = 1.7 \cdot {10}^{- 5}$

$x = \frac{1.7 \cdot {10}^{- 5}}{0.1} = 1.7 \cdot {10}^{- 4}$

The total concentration of the hydronium ions will thus be

$\left[{H}_{3} {O}^{+}\right] = 0.0100 + 0.00017 = \textcolor{g r e e n}{\text{0.01017 M}}$

As predicted, the concentration of the hydronium ions will largely be determined by the concentration of the strong acid.

SIDE NOTE Without the strong acid present, the concentration of the hydronium ions will be equal to

$\left[{H}_{3} {O}^{+}\right] = 1.3 \cdot {10}^{- 3} \text{M}$