# What will be the final pH on mixing 50 ml of 0.01 M methanoic acid with 50 ml of 0.01 M sodium hydroxide ?

## $p {K}_{a}$ for methanoic acid is 3.75.

May 4, 2017

$\textsf{p H = 8.22}$

#### Explanation:

$\textsf{H C O O H + N a O H \rightarrow H C O O N a + {H}_{2} O}$

$\textsf{{n}_{H C O O H} = c \times v = 0.10 \times \frac{50}{1000} = 0.005}$

$\therefore$sf(n_(HCOONa) which have been formed$\textsf{= 0.005}$

From the equation you can see that the total moles of NaOH added will also be 0.005.

The volume of 0.1M NaOH added must also be 50 ml at equivalence.

$\therefore$ Total volume = 50 + 50 = 100 ml = 0.1 L

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

You should be given the $\textsf{p {K}_{a}}$ value for $\textsf{H C O O H}$.

$\textsf{p {K}_{a} = 3.75}$

You need the $\textsf{p {K}_{b}}$ for $\textsf{H C O {O}^{-}}$.

You get this from:

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

$\therefore$$\textsf{p {K}_{b} = 14 - 3.75 = 10.25}$

Now use this expression for the pOH of a weak base:

sf(pOH=1/2(pK_b-log[base])

You get this from an ICE table and assume the dissociation is negligible.

$\therefore$sf(pOH=1/2(10.25-log[0.05])

$\textsf{p O H = \frac{1}{2} \left(10.25 - \left(- 1.3\right)\right) = 5.775}$

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

$\therefore$$\textsf{p H = 14 - p O H = 14 - 5.775 = 8.22}$

$\textsf{H C O O N a}$ is the salt of a weak acid and a strong base so you would expect the pH to be slightly alkali due to salt hydrolysis:

$\textsf{H C O {O}^{-} + {H}_{2} O \rightarrow H C O O H + O {H}^{-}}$