Question

Given: 50 mL of 0.1M ammonium nitrate (NH4NO3). Calculate the pH

Given: 50 mL of 0.167M NaC2H3O2. Calculate the pH

Answer 1

1) NH4NO3 is salt of weak base and strong acid

for such salts

pOH = 1/2 [pKb + pKw + log C]

C = 0.1 M

pKb of NH3 = 4.75

pOH = 1/2 [4.75 + 14 + log 0.1]

pOH = 1/2 [17.75]

pOH = 8.88

pH = 14 - 8.88

pH = 5.12

2) NaC2HO3 is salt of weak acid and strong base

for such salts

pH = 1/2 [pKw + pKa + log C]

pKa of CH3COOH = 4.75

C = 0.167 M

pH = 1/2 [14 + 4.75 + log 0.167]

pH = 9.0