A given volume of a buffer solution contains #6.85 x 10^-3# mol of the weak acid HY and #2.98 x 10^-3# mol of the salt NaY. The pH of the buffer solution is 3.78. How do you calculate the value of #pKa# for the acid HY at this temperature?

Question

A given volume of a buffer solution contains #6.85 x 10^-3# mol of the weak acid HY and #2.98 x 10^-3# mol of the salt NaY. The pH of the buffer solution is 3.78.  How do you calculate the value of #pKa# for the acid HY at this temperature?

Ellie Andrews · Accepted Answer

Well, the buffer equation holds that.... #pH=pK_a+log_10{[[Y^-]]/[[HY]]}#...

...we finally get #pK_a=3.42...# Now we were quoted molar quantities of the acid and its conjugate base....NOT concentrations...but given we have the quotient...#"concentration"="moles of solute"/"volume of solution"#...and in the buffer expression we get.... #log_10{((2.98.xx10^-3*mol)/(cancel"some volume"))/((6.85xx10^-3*mol)/cancel("some volume"))}#...that is the volumes cancel out....so WE DO NOT NEED TO KNOW THE VOLUME.... And so we solve for #pK_a# in the buffer equation.... #pK_a=pH-log_10{(2.98xx10^-3*mol)/(6.85xx10^-3*mol)}# #=3.78-underbrace(log_10{(2.98)/(6.85)})_"-0.361"=3.42# #pH_"solution"# is more acidic than #pK_a# given that the parent acid is present in GREATER concentration.....to what would #pH# be equal if #[HY]-=[Y^-]#?

Mateo Aguilar · Answer

undefined $$ \text{pKa} = \text{pH} + \log\left(\frac{\text{[A-]}}{\text{[HA]}}\right) $$
$$ \text{pKa} = 3.78 + \log\left(\frac{2.98 \times 10^{-3}}{6.85 \times 10^{-3}}\right) $$
$$ \text{pKa} \approx 3.78 + \log(0.435) $$
$$ \text{pKa} \approx 3.78 - 0.362 $$
$$ \text{pKa} \approx 3.418 $$