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?

1 Answer
Jun 27, 2018

Answer:

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

...we finally get #pK_a=3.42...#

Explanation:

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^-]#?