[chanelo]

So the question says to
a) calculate the change in pH when 5.00 mL of 0.100 mol/L HCl is added to a 100.0 mL buffer solution that is 0.100 mol/L in NH₃ and 0.100 mol/L NH₄Cl
b) calculate the change in pH when 5.00 mL of 0.100 mol/L NaOH is added to the original buffer.

* pKb of NH₃ is 4.74
I started by calculating the initial pH
pOH = pKb + log (0.100 mol/L NH₄⁺ / 0.100 mol/L NH₃)
pOH = 4.74
pH = 14.00 - pOH
pH = 9.26

mol NH₄⁺ = mol NH₃ = 0.100 L x 0.100 mol/L = 0.0100 mol
mol H⁺ added = 0.005 L x 0.100 mol/L = 0.0005 mol

mol NH₃ = 0.0100 mol - 0.0005 mol = 0.0095 mol
mol NH₄⁺ = 0.0100 mol + 0.0005 mol = 0.0105 mol

pOH = pKa + log (0.0105 / 0.0095)
pOH = 4.7835
pH = 9.2165

change in pH = 0.0435

I did something similar with NaOH and got a pH of 9.238 and the change as 0.018.
But both my answers were wrong so could someone please tell me where I went wrong?

[Borek]

In general you are on the right track. I would start simplifying things by converting pKb to pKa and using standard version of the Henderson-Hasselbalch equation (that is, pH=pKa+...)

pH change should be identical in both cases (just in a different direction). If you are asked to enter the difference, watch the sign and significant digits.

[chanelo]

Thanks, you were right! I just needed to add the "-" before 0.0435, and the second part was +0.0435.