buffer

(a)
NH3(aq) + H2O(l) = NH4+(aq) + OH-(aq)
Kb = 1.8 x 10-5 M
pKb = -log(1.8 x 10-5) = 4.74

pOH = pKb + log([NH4+]/[NH3]) = 4.74 + log(0.19/0.2) = 4.72
pH = 14 - pOH = 14 - 4.72 = 9.28

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(b)
Assume that total volume after mixing = 10 + 70 = 80 mL
HCl reacts with NH3(aq) to give NH4+(aq).
H+(aq) + NH3(aq) → NH4+(aq)

[NH3] = (0.2 x 70 - 0.1 x 10) / 80 = 0.1625 M
[NH4+] = (0.19 x 70 + 01 x 10) / 80 = 0.1788 M
pOH = pKb + log([NH4+]/[NH3]) = 4.74 + log(0.1788/0.1675) = 4.77
pH = 14 - pOH = 14 - 4.77 = 9.23