Question

a. What is the pH of a buffer solution that is 0.24 M NH₃ and 0.24 NH₄⁺? K_b for NH₃ is 1.8x10^-5.

b. What is the pH if 17 mL of 0.27 M hydrochloric acid is added to 515 mL of this buffer?

Answer 1

a) mixture of NH3 and NH4⁺ act as basic buffer

pOH = pKb + log [NH3⁺] / [NH3]

pKb = - log Kb = - log [1.8 x 10^-5] = 4.74

pOH = 4.74 + log [0.24] / [0.24]

pOH = 4.74

pH = 14 - 4.74

pH = 9.26

b) after HCl

initial millimoles of NH3 = 515 x 0.24 = 123.6

millimoles of NH4⁺ = 515 x 0.24 = 123.6

millimoles of HCl added = 17 x 0.27 = 4.59

after HCl added

millimoles of NH3 = 123.6 - 4.59 = 119.01

millimoles of NH4⁺ = 123.6 + 4.59 = 128.19  

totalvolume = 515 + 17 = 532 mL

[NH3] = 119.01 / 532 = 0.224 M

[NH4⁺] = 128.19 / 532 = 0.241 M

pOH = 4.74 + log [0.241] / [0.224]

pOH = 4.77

pH = 14 - 4.77

pH = 9.23