# A total of 100. mL of 0.1000M HClO (Ka= 3.0x10-8)is titrated with 0.0100M KOH at 25C?

## A: Calculate initial pH of the HClO solution B: What is the pH of the solution at the equivalence point, and how many mL of KOH solution must be added to reach the equivalence point?

Nov 17, 2016

$\textsf{\left(A\right)}$

$\textsf{p H = 4.26}$

$\textsf{\left(B\right)}$

$\textsf{p H = 9.74}$

$\textsf{1000 \textcolor{w h i t e}{x} m l}$ must be added.

#### Explanation:

$\textsf{\left(A\right)}$

$\textsf{H C l O}$ is a weak acid and dissociates:

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

For which:

$\textsf{{K}_{a} = \frac{\left[{H}^{+}\right] \left[C {l}^{-}\right]}{\left[H C l O\right]} = 3.0 \times {10}^{- 8} \textcolor{w h i t e}{x} \text{mol/l}}$

Since we have a weak acid we can rearrange and take -ve logs of both sides to get:

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

$\textsf{a}$ is the concentration of the acid which we can assume to be very close to the equilibrium concentration.

$\textsf{p {K}_{a} = - \log \left({K}_{a}\right) = - \log \left(3.0 \times {10}^{- 8}\right) = 7.522}$

$\therefore$$\textsf{p H = \frac{1}{2} \left[7.522 - \left(- 1\right)\right] = 4.26}$

$\textsf{\left(B\right)}$

The equation is:

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

This tells us that 1 mole of HClO is equivalent to 1 mole of KOH.

We can find the number of moles of HClO:

$\textsf{n = c \times v = 0.1000 \times \frac{100}{1000} = 0.01}$

The equation tells us that the no. moles of KOH must be the same:

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

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

$\therefore$$\textsf{v = \frac{c}{n} = \frac{0.01}{0.01} = 1 \text{L}}$

This is the volume of KOH solution required to reach the equivalence point.

This is a badly set question as you would never set up a titration that required such a large volume. The acid is far too concentrated.

It should be diluted 10 times and only 25 ml used. That would give a sensible end - point.

We have added 0.01 moles of HClO to 0.01 moles of KOH to make 0.01 moles KClO.

We can use a corresponding expression to get the pH of a weak base:

$\textsf{p O H = \frac{1}{2} \left[p {K}_{b} - \log b\right]}$

To get $\textsf{p {K}_{b}}$ we use:

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

$\therefore$$\textsf{p {K}_{b} = 14 - p {K}_{a} = 14 - 7.522 = 6.478}$

$\textsf{b = \left[C l {O}^{-}\right]}$

To get this we use $\textsf{c = \frac{n}{v}}$

We know that $\textsf{n = 0.01}$

The new total volume = 1000ml + 100ml = 1100 ml = 1.1 L

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

Putting in the numbers:

$\textsf{p O H = \frac{1}{2} \left[6.478 - \left(- 2.041\right)\right] = 4.2595}$

Now we can use:

$\textsf{p H + p O H = 14}$

$\therefore$$\textsf{p H = 14 - p O H = 14 - 4.2595 = 9.74}$