What is the final pH of the solution?

Question

Calculate the pH of the solution that results from the addition of 0.020 moles of HNO3 to a buffer made by combining
0.500 L of 0.060 M NH4Cl ( use Ka = 5.60 x 10 -10 )
and 0.500 L of 0.120 M.


The question is apparently cut off at the, but I believe you do not have to necessarily know the substance, since NH4Cl is a salt.

pKa=9.25181


Everytime I have done this problem, I have gotten an incorrect answer. Please help. I have been using the Henderson-Hasselbalch equation to solve this.

(-log(5.60 x (10^(-10)))) + log(.01 / .08) = 8.34872199

I don't understand what I'm doing wrong.

I thought that maybe if I switched it up (I don't know why), it would work but that is also wrong.

(-log(5.60 x (10^(-10)))) + log(.08 / .01) = 10.154902

Answer 1

First of all, you MUST have a full understanding of the problem before trying to attack it.
We have to necessarily know the substance. But as you said, since NH4Cl is a salt, and also since HCl is a strong acid, the other (missing) one must be NH4OH.
0.500 L of 0.120 M NH4OH <==> 0.06 mole NH4OH.
By adding 0.020 moles of HNO3 into the solution, the amount of NH4OH reduces to 0.04 mole and the amount of salt (considering to be NH4Cl) increases to 0.05 mole.
Further, assume that adding HNO3 did not increase the volume of the buffer, we must have: [NH4OH] = 0.02M and [NH4Cl] = 0.025M.
How do you get 0.01 and 0.08?
Also pay attention to Significant Figures!!