Sodium acetate is the salt of a strong base and a weak acid so we would expect it to be slightly alkali due to salt hydrolysis:
#sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^-)#
For which:
#sf(K_b=([CH_3COOH][OH^-])/([CH_3COO^-]))#
Before we set up an ICE table we need to find its initial concentration.
The number of moles is given by:
#sf(n_(CH_3COONa)=m/M_r=4.0/82.034=0.0487)#
We get the initial concentration using #sf(c=n/v)#:
#sf([CH_3COONa]=0.0487/0.500=0.09752color(white)(x)"mol/l")#
Now set up the ICE table based on #sf("mol/l")#:
#sf(" "CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^-)#
#sf(I" "0.09752" "0" "0)#
#sf(C" "-x" "+x" "+x)#
#sf(E" "(0.09752-x)" "x" "x)#
#:.##sf(K_b=(x^2)/((0.09752)-x))#
Since we are given #sf(pK_a)# we use this expression:
#sf(pK_a+pK_b=14)#
#:.##sf(pK_b=14-4.74=9.26)#
This means #sf(K_b)# is a very small number, so small that we can make the approximation that #sf((0.09752-x)rArr0.09752)#
So we can write:
#sf(K_b=x^2/0.09752)#
#:.##sf(x^2=K_bxx0.09752)#
#sf(x=sqrt(K_bxx0.09752)#
Since #sf(x=[OH^-])# we can take negative logs of both sides to get:
#sf(pOH=1/2[pK_b-log(0.09752)])#
#sf(pOH=1/2[9.26-(-1.01090)])#
#sf(pOH=5.1354)#
Since #sf(pH+pOH=14)# we get:
#sf(pH=14-pOH=14-5.1354=8.86)#
As expected, the pH is slightly alkaline.
