Chemistry pH problem?

if .040 moles of solid naOH is added to 1 liter of a solution that is .10 M in NH3 and .20 M in NH4Cl, what will be the pH of the resulting solution, assume no volume change due to addition of NaOH.

Answer:
let's set up a reaction:

NH4+ + OH- -->NH3 + H2O

now lets put in all the values.

NH4 + = 0.2M

NH3 = 0.1M

OH- = 0.04M

NH4+ + OH- -->NH3 + H2O

0.2M...0.04M...0.1M
-.04..-0.04..+0.04
0.16...0....0.14

i just subracted by the limiting factor and continued like a normal reaction chart.

now, to find the pH, we need to set up the buffer equation.

[OH-] = Kb x moles base/ moles acid

[OH-] = 1.8 x 10 ^-5 x 14/16 (or 0.14 / 0.16)

[OH-] = 1.575 x 10^ -5

take the log of that and add 14.

log (1.575 x 10^ -5) +14 = 9.20.
Hummm .040moles = molecules? to 1 liter WOW and to assume no volume change!?? You'd better becareful and sorry read or study! Assuming anything is the biggest no no !! Even in a controlled situation.

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