What is the pH of a buffer solution prepared from 4.17 g?

Question

What is the pH of a buffer solution prepared from 4.17 g of NH4NO3 (MM=80.04g/mol) and 0.20 L of 0.30 M NH3? What is the pH of the solution if diluted with pure water to 1000 mL? Kb = 1.8 x 10-5

Answer 1

Let NH4NO3, ammonium nitrate, be called AN (MW=80.0)
Let NH3 be written NH4OH

4.17gAN/0.2L x 1molAN/80gAN = 0.26 mol/L (before the dilution to 1000mL, assuming adding the AN does not change the volume)

4.17gAN/1L x 1molAN/80gAN = 5.21 x 10^-2 mol/L (after the dilution to 1000 mL)

NH4OH <===> NH4+ + OH-

[NH4+][OH-]/[NH4OH] = 1.85 x 10^-5

(0.26)[OH-]/(0.3) = 1.85 x 10^-5

[OH-] = (1.85 x 10^-5)(0.3)/(0.26) = 2.13 x 10^-5

pOH = -Log[OH-] = 4.67

pH + pOH = 14

pH = 9.33 (This is the first answer.)

The dilution to 1000 mL reduces [NH4OH] to 0.06M. We have already calculated the effect on the added NH4NO3.

(5.21 x 10^-2)[OH-]/(0.06) = 1.85 x 10^-5

[OH-] = (1.85 x 10^-5)(0.06)/(5.21 x 10^-2) = 2.13 x 10^-5

pOH = 4.67

pH = 9.33 (This is the second answer.)