The buffer equation tells us that......
Thus #pH=pK_a+log_10{[[A^-]]/[[HA]]}#
We gots #pK_a=-log_10(3.52xx10^-5)=4.45#, we have to work out the concentrations of free acid, and conjugate base.
Now addition of the strong base to the weak acid.....immediately gives us stoichiometric quantities of #Na^+A^-#, where #HA# was the conjugate acid........
#HA(aq) + NaOH(aq) rarr Na^+A^(-) + H_2O#
Starting moles of ..................#HA=0.615*Lxx0.250*mol*L^-1=0.154*mol# #(i)#
Moles of #NaOH=0.500*Lxx0.130*mol*L^-1=#
#0.0650*mol#.
And so INITIALLY we gots #(0.154-0.0650)*mol# #HA# #=# #0.089*mol# #(ii)#
And #0.0650*mol# #A^-#. These molar quantities are dissolved in a volume of #(500+615)*mL=1115*mL=1.115*L#, but it doesn't really matter because.....
#pH=pK_a+log_10{[[A^-]]/[[HA]]}#, and since in the log quotient....the volume of solution #"IS THE SAME"# we can ignore it, and just use the molar quantities of acid and conjugate base.
And so FINALLY.......
#pH=underbrace(-log_10(3.52xx10^-5))_color(red)(pK_a)+log_10{(0.0650)/(0.089)}#
#=# #-(-4.45)+(-0.136)=4.31......#
Now had #[A^-]=[HA]#, then the #pH# of the resultant buffer would have been equal to #pK_a#. Why because #log_(10)1=0#, and had we put this value back into the expression we get #pH=pK_a#.
The final #pH# is a trifle less than the #pK_a#, which makes sense given that there were more moles of acid than there were moles of base.....and the #pH# should be reduced from #pK_a# (because more #[H_3O^+]#.
I acknowledge that I have gone on for a long time about nothing to the power of less. But all I have done is to use #log# functions and calculated the molar quantities of acid and base, even tho it takes forever. Good luck.
