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.