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?

Answer 1

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^-]`?