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

Question

The $sf(pK_a)$ value for ethanoic acid is 4.66.

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Answer 1 · 2016-10-07T17:31:15

$sf(pH=8.68)$

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

$sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^(-))$

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

Note that these refer to equilibrium concentrations.

We can find the $sf(pOH)$ hence the $sf(pH)$ if we can get the value of $sf(pK_b)$ .

$sf(pK_a+pK_b=14)$ at $sf(25^@C)$

$:.$ $sf(pK_b=14-4.66=9.34)$

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 $sf(K_b)$ and taking negative logs of both sides we get:

$sf(pOH=1/2[pK_b-logb])$

Where $sf(b)$ is the concentration of the co - base.

Putting in the numbers:

$sf(pOH=1/2[9.34-log0.05]=1/2[9.34-(-1.3)]=5.32)$

$sf(pOH+pH=14)$

$:.$ $sf(pH=14-5.32=8.68)$