# What is the pH of a 0.05 M solution of sodium ethanoate ?

## The $\textsf{p {K}_{a}}$ value for ethanoic acid is 4.66.

Oct 7, 2016

$\textsf{p H = 8.68}$

#### Explanation:

Sodium ethanoate is the salt of a weak acid and strong base so undergoes hydrolysis:

$\textsf{C {H}_{3} C O {O}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s C {H}_{3} C O O H + O {H}^{-}}$

sf(K_b=([CH_3COOH][OH^(-)])/([CH_3COO^(-)])

Note that these refer to equilibrium concentrations.

We can find the $\textsf{p O H}$ hence the $\textsf{p H}$ if we can get the value of $\textsf{p {K}_{b}}$.

$\textsf{p {K}_{a} + p {K}_{b} = 14}$ at $\textsf{{25}^{\circ} C}$

$\therefore$$\textsf{p {K}_{b} = 14 - 4.66 = 9.34}$

Because the equilibrium is well to the left we can assume that the equilibrium concentration of sf(CH_3COO^(-) is very close to 0.05 M.

By rearranging the expression for $\textsf{{K}_{b}}$ and taking negative logs of both sides we get:

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

Where $\textsf{b}$ is the concentration of the co - base.

Putting in the numbers:

$\textsf{p O H = \frac{1}{2} \left[9.34 - \log 0.05\right] = \frac{1}{2} \left[9.34 - \left(- 1.3\right)\right] = 5.32}$

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

$\therefore$$\textsf{p H = 14 - 5.32 = 8.68}$