Question

Calculate the pH of 100 ml of a buffer that is .050 M NH4Cl and .160 M NH3 before and after the addition of 1.0 mL of 5.25 M HNO3

Before= 9.75

After?

Answer 1

pOH = pKb + log(NH4+/NH3)

pKb 0 4.75

then

initially

pOH = 4.75 + log(0.05/0.16) = 4.244

pH = 14-pOH = 14-4.244 = 9.756

then, after addition:

mmol of NH3 initially = MV = 100*0.16 = 16 mmol of NH3

mmol of NH4+ initially = MV = 100*0.05 = 5 mmol of NH3

after the addition of V = 1 ml of HNO3

mmol of acid added = MV = 1*5.25 = 5.25 mmol

then

mmol of NH3 after= 16-5.25 = 10.75

mmol of NH4+ after= 5+5.25 = 10.25

then

pOH = 4.75 + log(10.25/10.75) = 4.729

pH = 14-pOH = 14-4.729 = 9.271

slighlty more acidic than before