How do I solve this pH problem on NaOH

Question

If 0.040 moles of solid NaOH is added to 1.0L of a solution that is 0.10M in NH3, and 0.20M in NH4CL, what will be the oH of the resulting solution.

 

A. 4.80

B. 8.95

C. 5.05

D. 8.65

E. 9.20

Answer 1

This is another buffer problem where you have NH3 + NH4Cl making up the buffer.  Then NaOH is added.

When OH- is added it will react with the NH4+ to produce NH3 + H2O

moles of OH- added = 0.040

final [NH4+] = 0.20 - 0.04 = 0.16 M

final [NH3] = 0.10 + 0.04 = 0.14 M

pOH = pKb + log [salt]/[base] and for NH3 the pKb is 4.75 (look it up)

pOH = 4.75 + log (0.16/0.14) = 4.75 + 0.058

pOH = 4.81

pH = 14 - 4.81 = 9.19

So answer is probably E (9.20) again this will depend on the value used for pKb and rounding errors.