part A
A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 20.0mL of KOH.
part b
A 75.0-mL volume of 0.200 M NH3 (Kb=1.8
a) moles HBr = 0.0500 L x 0.15 M=0.0075
moles KOH = 0.0200 L x 0.25 M=0.0050
moles H+ in excess = 0.0075 - 0.0050= 0.0025
total volume = 0.070 L
[H+]= 0.0025/0.070=0.0357 M
pH =1.45
b) NH3 + HNO3 --> NH4NO3
0.075L x 0.2M NH3 = 0.015moles
0.015L x 0.5M HNO3 = 0.0075moles
HNO3 will react with NH3 in a 1:1 manner. therefore, 0.0075moles
NH3 will be 'removed' as NH4NO3 leaving 0.0075moles NH3 in 102ml
solution
0.0075moles / 0.102L = 0.0735M
1.8x10^-5 = [NH4+][OH-] / [NH3]
1.8x10^-5 = x^2 / 0.0735-x
x^2 + 1.8x10^-5x - 1.32x10^-6 = 0
x = [OH-] = 1.139x10^-3M
pH = 14 - pOH = 11.05
c.)
0.052L x 0.35M = 0.0182moles CH3COOH
0.4M x 0.015L = 0.0132 moles NaOH
NaOH + CH3COOH --> H2O + NaC2H3O2 1:1 ratio
therefore, all of the NaOH will be used up leaving 0.005moles
CH3COOH in 67ml
0.005moles / 0.067L = 0.075M CH3COOH
1.8x10^-5 = [H+][CH3COO-] / [CH3COOH]
1.8x10^-5 = x^2 / 0.075 - x
x^2 + 1.8x10^-5x - 1.35x10^-6 = 0
x = [H+] = 1.153 x 10^-3 M
pH = -log[H+] = 2.94
part A A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 20.0mL of KO...
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