Question

1) A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 11.0mL of KOH

2) A 75.0-mL volume of 0.200 M NH3 (Kb=1.8

Answer 1

1) moles of HBr = 0.15 X 50 X 10^-3 = 0.0075

Moles of KOH = 11 x 0.25 X 10^-3 = 0.00275

so moles of HBr neutralized = 0.00275

so moles of HBr left = 0.00475

molarity of HBr = moles / volume = 0.00475 X 1000 / (50+11) = 0.077M

so pH = -log (H+) = -log(0.077) = 1.11

2) Again we have

75 mL of 0.2 M NH3 which is equal to 75 X 0.2 M = 15 mmoles
add millimoles of H+ ( froma acid) = 21 mL x 0.5 M = 10.5 mmoles of H+

reaction

NH3 + H+ = NH4+

By reaction of 10.5 milllimoles of H+ it will give 10.5 millimoles of NH4+

this will leave behind 15-10.5 = 4.5 mmoles of the base

so concentration of NH4+ = 10.5 / 75+21 = 0.109M

and cocnentration of NH3 = 4.5 / 96 = 0.046M

from Hendersen equation we know that

pOH = pKb + log [NH4+]/ [NH3]= 4.7 + log 0.109/ 0.046=5.07

pH = 14 - pOH = 14 - 5.07 =8.93

3) As above

moles acetic acid = 0.0520 L x 0.35 M = 0.0182
moles OH- = 0.0250 L x 0.40 M=0.010
CH3COOH + OH- = CH3COO- + H2O
moles CH3COO- = 0.010
moles acetic acid in excess = 0.0182 - 0.010=0.0082
total volume = 52.0 + 25.0 = 77.0 mL = 0.0770 L
[acetic acid] = 0.0082/ 0.0770=0.106 M
[acetate]= 0.010/ 0.0770 = 0.129 M
pKa = 4.7
pH = 4.7 + log 0.129/ 0.106 =4.785