We interrogate the equilibrium....
#C_6H_5CO_2H(aq) + H_2O(l) rightleftharpoonsC_6H_5CO_2^(-)+H_3O^+#
And as usual...#K_a=([C_6H_5CO_2^(-)][H_3O^+])/([C_6H_5CO_2H])=10^(-4.202)=6.28xx10^-5#...data from this site..
And so if #x*mol*L^-1# benzoic acid dissociates then...
#6.28xx10^-5=x^2/((10.0*g)/(122.12*g*mol^-1)-x)=x^2/(0.0819-x)#
...a quadratic in #x# that could be solved EXACTLY...we make the approx. that #0.0819">>"x#...and get a first approx. that...
#x_1=sqrt(6.28xx10^-5xx0.0819)=0.00227*mol*L^-1#...the which is small compared to #0.0819#...but we can pursue successive approximation....
#x_2=sqrt(6.28xx10^-5xx(0.0819-0.00227))=0.0224*mol*L^-1#
#x_3=sqrt(6.28xx10^-5xx(0.0819-0.00224))=0.0224*mol*L^-1#
I am prepared accept this value as close to the true one...
And so #[H_3O^+]-=[C_6H_5CO_2^(-)]-=0.0224*mol*L^-1#...#pH=-log_10(0.0224)=+2.65#...
