Question

Part B

A 72.0 mL volume of 0.25 M  HBr is titrated with 0.50 M  KOH. Calculate the pH after addition of 36.0 mL of KOH at 25 ∘C.

Express the pH numerically.

Part C

Consider the titration of 50.0 mL of 0.20 M  NH3 (Kb=1.8×10−5) with 0.20 M  HNO3. Calculate the pH after addition of 50.0 mL of the titrant at 25 ∘C.

Express the pH numerically.

Part D

A 30.0-mL volume of 0.50 M  CH3COOH (Ka=1.8×10−5) was titrated with 0.50 M  NaOH. Calculate the pHafter addition of 30.0 mL of NaOH at 25 ∘C.

Express the pH numerically.

Answer 1

1) for this titration , we have to calculate moles of HBr and KOH ,

moles of HBr = 0.072 L x 0.25 M=0.018 mol

moles of KOH = 0.036 L x 0.50 M=0.018 mol

we know that HBr is strong acid and KOH is strong base

therefore pH = 7.0

2) we have to calculate moles of NH3 and HNO3, therefore

moles of NH3 = 0.050 L x 0.20 M=0.010 mol

moles of HNO3 = 0.050 L x 0.20 M= 0.010 mol

the reaction is given as

NH3 + H+ ------------> NH4+

moles of NH4+ = 0.010 mol

total volume of the solution = 50.0 ml + 50.0 ml = 100.0 mL = 0.100 L

therefore [NH4+]= 0.010 mol / 0.100 L = 0.100 mol/L

the dissociation of NH4+ is given as

NH4+ + H2O <------> NH3 + H3O+

Ka = Kw/Kb = 1.0 x 10:14 / 1.8 x 10^-5 = 5.6 x 10^-10 = x^2 / (0.10M-x )

x =7.45 x 10^-6 M

pH = 5.13

3) first we have to calculate moles of CH3COOH and NaOH

moles of acetic acid = 0.030 L x 0.50 M = 0..015 mol

moles of NaOH = 0.030 L x 0.50 M = 0.015 mol

total volume = 30 ml + 30 ml = 0.060 L

the reaction of CH3COOH and NaOH is given as

CH3COOH + OH- = CH3COO- + H2O

moles of acetate = 0.015 mol / 0.060 L= 0.25 mol/L

then CH3COO- reacts with H2O , and we get

CH3COO- + H2O <----> CH3COOH + OH-

Kb = Kw/Ka = 5.6 x 10^-10 = x^2/ 0.25-x

x = [OH-]=1.2 x 10^-5 M

pOH = 4.9

pH = 14 - 4.9 = 9.1