# Question #fc66b

Apr 20, 2017

$\textsf{p H = 4.76}$

#### Explanation:

Ethanoic acid is a weak acid and dissociates:

$\textsf{C {H}_{3} C O O H r i g h t \le f t h a r p \infty n s C {H}_{3} C O {O}^{-} + {H}^{+}}$

For which:

$\textsf{{K}_{a} = \frac{\left[C {H}_{3} C O {O}^{-}\right] \left[{H}^{+}\right]}{\left[C {H}_{3} C O O H\right]} = 1.74 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

Remember these refer to equilibrium concentrations.

You would normally be given this value in a question.

To find the pH we need to find the $\textsf{{H}^{+}}$ concentration.

Rearranging gives:

$\textsf{\left[{H}^{+}\right] = {K}_{a} \times \frac{\left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]}}$

Because the value of $\textsf{{K}_{a}}$ is small we can make the important assumption that the initial concentrations given are a good enough approximation to the equilibrium concentrations.

Please note that the solutions have been mixed so have been diluted. A common error is to use these concentrations which you have been given in the calculation.

Perhaps this is where you were going wrong. We need to find the number of moles of acid and salt.

We know that $\textsf{c = \frac{n}{v}}$

So $\textsf{n = c \times v}$

So the number of moles of $\textsf{C {H}_{3} C O O H}$ is given by:

$\textsf{{n}_{C {H}_{3} C O O H} = c \times v = 0.02 \times 0.10 = 0.002}$

(Note that 100 ml is 0.10 L)

The new total volume is 100 + 50 = 150 ml = 0.150 L

$\therefore$$\textsf{\left[C {H}_{3} C O O H\right] = \frac{n}{v} = \frac{0.002}{0.150} \textcolor{w h i t e}{x} \text{mol/l}}$

The number of moles of $\textsf{C {H}_{3} C O {O}^{-}}$ is given by:

$\textsf{{n}_{C {H}_{3} C O {O}^{-}} = c \times v = 0.04 \times 0.05 = 0.002}$

$\therefore$$\textsf{\left[C {H}_{3} C O {O}^{-}\right] = \frac{n}{v} = \frac{0.002}{0.150} \textcolor{w h i t e}{x} \text{mol/l}}$

Now we can put these numbers into the expression for $\textsf{{H}^{+} \Rightarrow}$

$\textsf{\left[{H}^{+}\right] = 1.74 \times {10}^{- 5} \times \frac{\cancel{\frac{0.002}{0.150}}}{\cancel{\frac{0.002}{0.150}}}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left[1.74 \times {10}^{5}\right] = 4.76}$