The buffer equation tells us that......
Thus pH=pK_a+log_10{[[A^-]]/[[HA]]}pH=pKa+log10{[A−][HA]}
We gots pK_a=-log_10(3.52xx10^-5)=4.45pKa=−log10(3.52×10−5)=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^-Na+A−, where HAHA was the conjugate acid........
HA(aq) + NaOH(aq) rarr Na^+A^(-) + H_2OHA(aq)+NaOH(aq)→Na+A−+H2O
Starting moles of ..................HA=0.615*Lxx0.250*mol*L^-1=0.154*molHA=0.615⋅L×0.250⋅mol⋅L−1=0.154⋅mol (i)(i)
Moles of NaOH=0.500*Lxx0.130*mol*L^-1=NaOH=0.500⋅L×0.130⋅mol⋅L−1=
0.0650*mol0.0650⋅mol.
And so INITIALLY we gots (0.154-0.0650)*mol(0.154−0.0650)⋅mol HAHA == 0.089*mol0.089⋅mol (ii)(ii)
And 0.0650*mol0.0650⋅mol A^-A−. These molar quantities are dissolved in a volume of (500+615)*mL=1115*mL=1.115*L(500+615)⋅mL=1115⋅mL=1.115⋅L, but it doesn't really matter because.....
pH=pK_a+log_10{[[A^-]]/[[HA]]}pH=pKa+log10{[A−][HA]}, and since in the log quotient....the volume of solution "IS THE SAME"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.
