Question

The $pH$ of $0.02M$ solution of an unknown weak acid is $3.7$ , how would you find the $pKa$ of this acid?

Answer 1

Hint : We use $pH$ scale to know more about the acidity or basicity of given substance. In general, substance having $pH$ scale greater than $7$ is basic in nature and less than $7$ is acidic in nature. $pKa$ is the term we used to tell the strength of any acid. $pKa$ is a negative log of $Ka$ which is acid dissociation constant.

Complete Step By Step Answer:
We know $pH$ $= 3.7$
$- \log \left[ {{H^ + }} \right] = 3.7$
Taking antilog, $\left[ {{H^ + }} \right] = {10^{ - 3.7}}M$
As we know weak acid dissociates into its conjugate base and a proton, the moles of conjugate base is equal to moles of protons.
$HA \rightleftharpoons {A^ - } + {H^ + }$
That is why $\left[ {{A^ - }} \right] = \left[ {{H^ + }} \right]$
At equilibrium, the acid dissociation constant will be
$Ka = \dfrac{{{{\left[ {{H^ + }} \right]}_{eq}}^{{2_{}}}}}{{{{\left[ {HA} \right]}_{eq}}}}$
$HA$ gives away the proton and that is why it becomes ${A^ - }$ which is then subtracted from the moles of $HA$
Ka = \dfrac{{{{\left[ {{H^ + }} \right]}_{eq}}^2}}{{\left[ {HA} \right] - {{\left[ {{H^ + }} \right]}_{eq}}}} \\\ = \dfrac{{{{({{10}^{ - 3.7}})}^2}}}{{0.02 - {{10}^{ - 3.7}}}} \\\ = 2.01 \times {10^{ - 6}}M \\\
Now we have $Ka$ , we can find out $pKa$ ,
pKa = - \log Ka \\\ = - \log (2.01 \times {10^{ - 6}}) \\\ pKa \approx 5.7 \\\

Note :
The higher the dissociation constant the greater the strength of acid and it will have a high degree of dissociation. Weak acids have low dissociation constant value that means the acid will easily get ionised in the solution.