Explanation:
It is known that
of
=
.
(a) Relation between
and
is as follows.

Putting the values into the above formula as follows.

= 
= 3.347
Also, relation between pH and
is as follows.
pH = ![pK_{a} + log\frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
= 
= 3.44
Therefore, pH of the buffer is 3.44.
(b) No. of moles of HCl added = 
=
= 0.0116 mol
In the given reaction,
will react with
to form 
Hence, before the reaction:
No. of moles of
= 
= 0.15 mol
And, no. of moles of
= 
= 0.12 mol
On the other hand, after the reaction :
No. of moles of
= moles present initially - moles added
= (0.15 - 0.0116) mol
= 0.1384 mol
Moles of
= moles present initially + moles added
= (0.12 + 0.0116) mol
= 0.1316 mol
As,
= 

= 
= 3.347
Since, volume is both in numerator and denominator, we can use mol instead of concentration.
As, pH = ![pK_{a} + log \frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%20%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
= 3.347+ log {0.1384/0.1316}
= 3.369
= 3.37 (approx)
Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.