# A 0.500M solution of a weak acid, HX, has a ka=3.26x10^-5 a) What is the pH of this solution? b) What is the percent ionization? c) What would be the pH be in a solution containing the strong electrolyte, 0.2M AlX3

## A 0.500M solution of a weak acid, HX, has a ka=3.26x10^-5 a) What is the pH of this solution? b) What is the percent ionization? c) What would be the pH be in a solution containing the strong electrolyte, 0.2M AlX3? d) What is the percent ionization in part C? e) Does the presence of a strong electrolyte with a common ion increase or decrease the pH?

Jun 11, 2016

#### Answer:

$\left(a\right) \text{ } 2.4$

(b)" " 0.8%

$\left(c\right) \text{ } 4.56$

(d)" "0.0054%

$\left(e\right) \text{ ""increase}$

#### Explanation:

$\left(a\right)$

To find the pH we can set up an $\text{ICE}$ table:

$\text{ "HX_((aq))" "rightleftharpoons" "H_((aq))^(+)" "+" } {X}_{\left(a q\right)}^{-}$

color(red)("I")" "0.5" "0" "0

color(red)("C")" " -x" "+x" "+x

color(red)("E")" "(0.5-x)" "x" "x

The expression for ${K}_{a}$ is:

${K}_{a} = \frac{\left[{H}_{\left(a q\right)}^{+}\right] \left[{X}_{\left(a q\right)}^{-}\right]}{\left[H {X}_{\left(a q\right)}\right]}$

$\therefore {K}_{a} = {x}^{2} / \left(0.5 - x\right) = 3.26 \times {10}^{- 5} \text{ ""mol/l}$

Because of the small value of ${K}_{a}$ we can assume that $x$ is much smaller than $0.5$ so we can assume $\left(0.5 - x\right) \Rightarrow 0.5$

$\therefore {K}_{a} = {x}^{2} / 0.5 = 3.26 \times {10}^{-} 5$

$\therefore {x}^{2} = 3.26 \times {10}^{- 5} \times 0.5 = 1.63 \times {10}^{- 5}$

$\therefore x = \sqrt{1.63 \times {10}^{- 5}} = 4.037 \times {10}^{- 3} \text{ ""mol/l}$

$\therefore \left[{H}_{\left(a q\right)}^{+}\right] = 4.037 \times {10}^{- 3} \text{ ""mol/l}$

$p H = - \log \left[{H}_{\left(a q\right)}^{+}\right] = - \log \left(4.037 \times {10}^{- 3}\right)$

$\textcolor{red}{p H = 2.4}$

$\left(b\right)$

From an initial moles of $0.5$ only $4.037 \times {10}^{- 3}$ are ions so the percent ionisation =(4.037xx10^(-3))/(0.5)xx100=color(red)(0.8%)

$\left(c\right)$

Now the solution contains $0.2 \text{M}$ of $A l {X}_{3}$. As it is a strong electrolyte we can say that this a $0.6 \text{M}$ solution of ${X}_{\left(a q\right)}^{-}$

Now the $\text{ICE}$ table looks like this:

$\text{ "HX_((aq))" "rightleftharpoons" "H_((aq))^(+)" "+" } {X}_{\left(a q\right)}^{-}$

color(red)("I")" "0.5" "0" "0.6

color(red)("C")" " -x" "+x" "+x

color(red)("E")" "(0.5-x)" "x" "(0.6+x)

$\therefore {K}_{a} = \frac{x \left(0.6 + x\right)}{0.5 - x}$

Again, because $x$ is much smaller than $0.5$ and $0.6$ this simplifies to:

${K}_{a} = \frac{0.6 x}{0.5}$

$\therefore {K}_{a} = 1.2 x = 3.26 \times {10}^{- 5}$

$\therefore x = \frac{3.26 \times {10}^{- 5}}{1.2} = 2.716 \times {10}^{- 5} = \left[{H}_{\left(a q\right)}^{+}\right]$

$p H = - \log \left[{H}_{\left(a q\right)}^{+}\right] = - \log \left(2.716 \times {10}^{- 5}\right)$

$\textcolor{red}{p H = 4.56}$

$\left(d\right)$

Now the percentage ionisation becomes:

(2.716xx10^(-5))/(0.5)xx100=color(red)(0.0054%)

$\left(e\right)$

You can see that adding extra ${X}_{\left(a q\right)}^{-}$ has raised the $p H$.

This is an example of "The Common Ion Effect" which is a consequence of Le Chatelier's Principle.

Adding a large amount of ${X}_{\left(a q\right)}^{-}$ has caused the position of equilibria to shift to the left, reducing the amount of ${H}_{\left(a q\right)}^{+}$ ions causing an increase in $p H$.