I'll derive the expression first.
Let acetic acid be HA:
It is a weak acid and dissociates:
#sf(HXrightleftharpoonsH^++X^-)#
For which:
#sf(K_a=[[H^+][X^-]]/[[HX]]=1.76xx10^(-5)color(white)(x)"mol/l")#
Let the initial concentration be #sf(Ccolor(white)(x)"mol/l")#.
Set up an ICE table in mol/l:
#sf(" "HX" "rightleftharpoons" "H^++X^-)#
#sf(I" "C" "0" "0)#
#sf(C" "-x" "+x" "+x)#
#sf(E" "(C-x)" "x" "x)#
#:.##sf(K_a=x^2/((C-x)))#
If #sf(K_a)# lies between about #sf(10^-4)# to #sf(10^(-13)# then we can assume that #sf(x)# is small enough to let #sf((C-x)rArrC)#.
#:.##sf(K_a=x^2/C)#
#:.##sf(x^2=[H^+]^2=K_aC)#
#:.##sf([H^+]=sqrt(K_aC)#
#sf([H^+]=sqrt(1.76xx10^(-5)xx3xx10^(-3))#
#sf([H^+]=+-2.297xx10^(-4)color(white)(x)"mol/l")#
Since the -ve root is undefined we get:
#sf(pH=-log[H^+]=-log(2.297xx10^(-4))=3.64)#
If you are given the #sf(pK_a)# value I'll show how to get another useful expression:
#sf([H^+]=sqrt(K_aC)=(K_aC)^(1/2))#
Taking logs of both sides:
#sf(log[H^+]=1/2logK_a+1/2logC)#
Multiply through by -1 #rArr#
#sf(-log[H^+]=-1/2logK_a-1/2logC)#
This becomes:
#sf(color(red)(pH=1/2[pK_a-logC]))#
This is quick way of finding the pH of a weak acid given the concentration.
