# PH Question?

Sep 19, 2016

(A)

pH = 4.26

(B)

pH = 9.7

V = 1 L

#### Explanation:

(A)

A quick way of finding the pH of a weak acid is to use the expression:

$\textsf{p H = \frac{1}{2} \left(p {K}_{a} - \log a\right)}$

$\textsf{p {K}_{a} = - \log {K}_{a} = - \log \left(3 \times {10}^{- 8}\right) = 7.523}$

$\textsf{a}$ is the concentration of the acid.

$\therefore$$\textsf{p H = \frac{1}{2} \left[7.523 - \left(- 1\right)\right] = 4.26}$

(B)

You need to do the volume first:

The equation is:

$\textsf{H C l O + K O H \rightarrow K C l O + {H}_{2} O}$

$\textsf{{n}_{H C l O} = c \times v = 0.1000 \times \frac{100}{1000} = 0.01}$

Since they react in a 1:1 molar ratio, the number of moles of base must be the same:

$\textsf{{n}_{K O H} = 0.01}$

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{v = \frac{n}{c} = \frac{0.01}{0.01} = 1 \textcolor{w h i t e}{x} L}$

This is a bad question as you would never set up a titration which requires 1L for an end-point.

$\textsf{C l {O}^{-}}$ is the co - base and is hydrolysed by water:

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

sf(K_(b)=([HClO][OH^(-)])/([ClO^(-)])

These are equilibrium concentrations.

We can use this expression to find $\textsf{\left[O {H}^{-}\right]}$ hence $\textsf{\left[{H}^{+}\right]}$ and the pH.

We can find $\textsf{{K}_{b}}$ using the expression:

$\textsf{{K}_{b} \times {K}_{a} = {K}_{w} = {10}^{- 14} \textcolor{w h i t e}{x} {\text{mol"^2."l}}^{- 2}}$ at $\textsf{{25}^{\circ} C}$

$\therefore$$\textsf{{K}_{b} = {10}^{- 14} / \left(3.0 \times {10}^{- 8}\right) = 0.333 \times {10}^{- 6} \textcolor{w h i t e}{x} \text{mol/l}}$

Now we can set up an ICE table based on concentrations in mol/l:

The total volume = 100 ml + 1000 ml = 1.1 L

$\therefore$ $\textsf{\left[C l {O}^{-}\right] = \frac{n}{v} = \frac{0.01}{1.1} = 0.00909 \textcolor{w h i t e}{x} \text{mol/l}}$

$\text{ }$$\textsf{C l {O}^{-} \text{ "+" "H_2O" "rightleftharpoons" "HClO" "+" } O {H}^{-}}$

$\textsf{\textcolor{red}{I} \text{ "0.00909" "0" } 0}$

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

$\textsf{\textcolor{red}{E} \text{ "(0.00909-x)" "x" } x}$

$\therefore$$\textsf{{K}_{b} = {x}^{2} / \left(\left(0.00909 - x\right)\right) = 0.333 \times {10}^{- 6} \textcolor{w h i t e}{x} \text{mol/l}}$

Because the dissociation is so small I will make the assumption that $\textsf{\left(0.00909 - x\right) \Rightarrow 0.00909}$.

$\therefore$$\textsf{{x}^{2} = 0.333 \times {10}^{- 6} \times 0.00909 = 0.0032 \times {10}^{- 6}}$

$\therefore$$\textsf{x = \sqrt{0.0032 \times {10}^{- 6}} = 5.5 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

This is equal to $\textsf{\left[O {H}^{-}\right]}$.

The ionic product of water is given by:

$\textsf{{K}_{w} = \left[{H}^{+}\right] \left[O {H}^{-}\right] = {10}^{- 14} \textcolor{w h i t e}{x} {\text{mol"^2."l}}^{-} 2}$ at $\textsf{{25}^{\circ} C}$

$\therefore$$\textsf{\left[{H}^{+}\right] = {10}^{- 14} / \left[\left[O {H}^{-}\right]\right] = {10}^{- 14} / \left(5.5 \times {10}^{- 5}\right) = 1.818 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left[1.818 \times {10}^{- 10}\right] = 9.7}$

This is the salt of a weak acid and a strong base so you can see that, at the equivalence point, the solution is slightly alkaline.