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

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

1 Answer
Oct 7, 2016

Answer:

#sf(pH=8.68)#

Explanation:

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)#