Explanation:
According to the Henderson-Hasselbalch equation for the solution before addition of HCl.
pH = ![pK_{a} + log(\frac{[A^{-}]}{[HA]}) ](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%29%0A)
Now, putting the given values into the above formula as follows.
8.0 = ![7.4 + log(\frac{[A^{-}]}{[HA]}) ](https://tex.z-dn.net/?f=7.4%20%2B%20log%28%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%29%0A)
= 4.0 ............ (1)
Also,
= 0.1 moles is the total of the compound.
Therefore,
= 0.1
= 0.1 - [HA] ............. (2)
Now, substituting the value of
from equation (2) into equation (1) as follows.
= 4.0
= 4.0
0.1 - [HA] = 4.0[HA]
0.1 = 5.0[HA]
[HA] = 0.02 mol
Therefore,
= 0.1 - [HA]
= 0.1 - 0.02
= 0.08 mol
Now, addition of strong acid reduces
and increases AH by the same amount.
(0.03 L) × (1 mol/L) = 0.03 moles HCl are added.
So,
= 0.08 - 0.03 = 0.05 moles
and, AH = 0.02 + 0.03 = 0.05 moles
Therefore, after HCl addition,
= 
= 1.0
Now, re-substituting into the Henderson-Hasselbalch equation:
pH = 7.4 + log(1.0)
= 7.4
Thus, we can conclude that pH of the resulting solution is 7.4.