Question

Calculate the pH change produced when 0.100 mol of gaseous HCl is added to each of the following buffer solutions.

a) 500 mL of 0.900 M NH3 and 0.900 M NH4Cl

b) 500 mL of 0.200 M NH3 and 0.800 M NH4Cl

c) 500 mL of 0.100 M NH3 and 0.900 M NH4Cl

Answer 1

pOH = pKb + log [acid]/[base]

pH + pOH = 14

a)

0.500L x 0.900M NH₄Cl = 0.450 mol NH₄⁺
0.500L x 0.900M NH₃ = 0.450 NH₃

0.450 + 0.100 = 0.550 mol NH₄⁺
0.450 - 0.100 = 0.350 mol NH₃

0.55 / 0.500 = 1.1M

0.35 / 0.500 =0.70M

K_b of NH₃ = 1.8 x 10^-5

pOH = 4.74 + log (1.1/.07)

pOH = 4.936

therefore;
pH = 14 -4.936
pH = 9.06

b)

Similarly;

0.500L x 0.800M NH₄Cl = 0.400 mol NH₄⁺
0.500L x 0.200M NH₃ = 0.100 NH₃
0.400 + 0.100 = 0.500 mol NH₄⁺
0.100 - 0.100 = 0 mol NH₃

pOH =4.74; pH = 9.26

c)

0.500L x 0.900M NH₄Cl = 0.450 mol NH₄⁺
0.500L x 0.100M NH₃ = 0.050 NH₃

0.450 + 0.100 = 0.550 mol NH₄⁺
0.050 - 0.100 = -0.050 mol NH₃

0.55 / 0.500 = 1.1M

-0.05 / 0.500 =-0.10M

pOH = 4.74 + log (1.1/-0.1) = 4.74-1.041 = 3.699

pH = 14-3.699 = 10.3