Question

1. What is the pH of a buffer solution that is .24M NH₃ and .20M NH₄Cl?

If .005 mol of NaOH is added to .50L of the buffer solution from question 1, what is the resulting pH?

If .03 mol of HCl is added to .05L of the buffer solution from question 1, what is the resulting pH?

Answer 1

1) mixture of NH3 and NH4Cl act as basic buffer

pOH = pKb + log [NH4Cl] / [NH3]

pKb = 4.75 for NH3

pOH = 4.75 + log [0.20] / [0.24]

pOH = 4.67

pH = 14 - 4.67

pH = 9.33

2) if we add 0.005/0.5 = 0.01 M NaOH added

[NH3] = 0.24 + 0.01 = 0.25 M

[NH4⁺] = 0.20 - 0.01 = 0.19 M

pOH = 4.75 + log [0.19] / [0.25]

pOH = 4.63

pH = 14 - 4.63

pH = 9.37

3) if we add 0.03 / 0.5 = 0.06 M HCl

[NH3] = 0.24 - 0.06 = 0.18 M

[NH3⁺] = 0.20 + 0.06 = 0.26 M

pOH = 4.75 + log [0.26] / [0.18]

pOH = 4.91

pH = 14 - 4.91

pH = 9.09