# 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?

Jun 27, 2018

Well, the buffer equation holds that.... $p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{Y}^{-}\right]}{\left[H Y\right]}\right\}$...

...we finally get $p {K}_{a} = 3.42 \ldots$

#### Explanation:

Now we were quoted molar quantities of the acid and its conjugate base....NOT concentrations...but given we have the quotient...$\text{concentration"="moles of solute"/"volume of solution}$...and in the buffer expression we get....

${\log}_{10} \left\{\left(\frac{2.98 . \times {10}^{-} 3 \cdot m o l}{\cancel{\text{some volume"))/((6.85xx10^-3*mol)/cancel("some volume}}}\right)\right\}$...that is the volumes cancel out....so WE DO NOT NEED TO KNOW THE VOLUME....

And so we solve for $p {K}_{a}$ in the buffer equation....

$p {K}_{a} = p H - {\log}_{10} \left\{\frac{2.98 \times {10}^{-} 3 \cdot m o l}{6.85 \times {10}^{-} 3 \cdot m o l}\right\}$

$= 3.78 - {\underbrace{{\log}_{10} \left\{\frac{2.98}{6.85}\right\}}}_{\text{-0.361}} = 3.42$

$p {H}_{\text{solution}}$ is more acidic than $p {K}_{a}$ given that the parent acid is present in GREATER concentration.....to what would $p H$ be equal if $\left[H Y\right] \equiv \left[{Y}^{-}\right]$?