The answer is God because if you think on it hard enough, you realize that we couldnt of come from nowhere. Then, you get a headache and just stop thinking about.
Hope this helps :D
This equation is impossible. NaSO4 is non-existent. Did you mean Na2SO4?
Answer:
A) pH of the buffer is 3.44
B) pH of the buffer solution is 3.37
Explanation:
Relation between K_{a} and pK_{a} is as follows.

.

The relation between pH and pK_{a} is as follows.
![pH = pK_{a} + log\frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pK_%7Ba%7D%20%2B%20log%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)

pH of the buffer is 3.44.
b)
mol of HCl added = 11.6M *0.001 L = 0.0116 mol
In the given reaction,
will react with
to form
No. of moles of

And, no. of moles of 
= 0.12 mol
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)
= 0.1316 mol
As,

Since, volume is both in numerator and denominator, we can use mol instead of concentration.
![pH = pK_{a} + log \frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pK_%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
pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37