# 2.5x10-3 moles of solid Mg(OH)2 is added to a 100.0 mL sample of 0.250 M aqueous chlorous acid solution. Ignore the volume change associated with the added solid. Find the pH of the solution?

## Use the information below to answer the following question: 2.5x10-3 moles of solid Mg(OH)2 is added to a 100.0 mL sample of 0.250 M aqueous chlorous acid solution. Ignore the volume change associated with the added solid. Find the pH of the solution.

Sep 16, 2016

$\textsf{p H = 1.6}$

#### Explanation:

The base is neutralised by the acid:

$\textsf{M g {\left(O H\right)}_{2} + 2 H C l {O}_{2} \rightarrow M g {\left(C l {O}_{2}\right)}_{2} + 2 {H}_{2} O}$

The number of moles of chlorous acid is given by:

$\textsf{{n}_{H C l {O}_{2}} = c \times v = 0.250 \times \frac{100.0}{1000} = 0.0250}$

We are told that:

$\textsf{{n}_{M g {\left(O H\right)}_{2}} = 0.0025}$

From the stoichiometry of the reaction we can see that the number of moles of $\textsf{H C l {O}_{2}}$ consumed must be $\textsf{0.0025 \times 2 = 0.005}$.

So the number remaining is given by:

$\textsf{{n}_{H C l {O}_{2}} = 0.025 - 0.005 = 0.02}$

We can also say that the number of moles of $\textsf{C l {O}_{2}^{-}}$ formed = $\textsf{0.0025 \times 2 = 0.005}$

We have now created an acid buffer since we have a weak acid in combination with its co - base.

Chlorous acid dissociates:

$\textsf{H C l {O}_{2} r i g h t \le f t h a r p \infty n s {H}^{+} + C l {O}_{2}^{-}}$

For which:

$\textsf{{K}_{a} = \frac{\left[{H}^{+}\right] \left[C l {O}_{2}^{-}\right]}{\left[H C l {O}_{2}\right]} = {10}^{- 12} \textcolor{w h i t e}{x} \text{mol/l}}$

The concentrations are those at equilibrium.

To find $\textsf{\left[{H}^{+}\right]}$ and hence the pH we need to set up an ICE table based on concentration.

Since we are told the volume is 100 ml = 0.1 L then the initial concentrations become:

sf([HClO_2]=0.02/0.1=0.2color(white)(x)"mol/l"

$\textsf{\left[H C l {O}_{2}^{-}\right] = \frac{0.005}{0.1} = 0.05 \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{\left[{H}^{+}\right] = 0 \textcolor{w h i t e}{x} \text{mol/l}}$

""" "sf(HClO_2" "rightleftharpoons" "H^(+)" "+" "ClO_2^-)

$\textsf{\textcolor{red}{I} \text{ "0.2" "0" } 0.05}$

$\textsf{\textcolor{red}{C} \text{ "-x" "+x" } + x}$

$\textsf{\textcolor{red}{E} \text{ "(0.2-x)" "x" } \left(0.05 + x\right)}$

$\therefore$sf(K_a=(x(0.05+x))/((0.2-x))=10^(-2)

At this point it is common to assume that, because the dissociation is so small then $\textsf{\left(0.05 + x\right) \Rightarrow 0.05}$ and $\textsf{\left(0.2 - x\right) \Rightarrow 0.2}$.

However, we can't do this in this case as the size of $\textsf{{K}_{a}}$ means that $\textsf{x}$ is significant.

If we multiply out the expression we get:

$\textsf{{x}^{2} + 0.06 x - 0.002 = 0}$

This is a quadratic equation which can be solved for $\textsf{x}$ using the quadratic formula. I won't go into this here but this gives, ignoring the absurd root:

$\textsf{x = 0.02385 \textcolor{w h i t e}{x} \text{mol/l} = \left[{H}^{+}\right]}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left(0.02385\right) = 1.6}$

If we make the assumption that $\textsf{\left(0.05 + x\right) \Rightarrow 0.05}$ and $\textsf{\left(0.2 - x\right) \Rightarrow 0.2}$ then $\textsf{x \frac{0.05}{0.2} = 0.01}$ which gives $\textsf{x = 0.4}$ for which $\textsf{p H = 1.4}$ which is significantly lower.