The Kb for ammonia is 1.8 X 10⁻⁵
The ammonia will dissociate as
NH₃ + H₂O ---> NH₄⁺ + OH⁻
initial 0.350 0 0
Change -x +x +x
Equilibrium 0.35-x x x
[tex]Kb=\frac{[NH_{4}^{+}[OH^{-} ]}{[NH_{3}]}[/tex]
Putting values
[tex]Kb=\frac{[x][x]}{[0.35-x]}=\frac{x^{2} }{0.35-x}[/tex]
We may ignore x in denominator as Kb is very low
therefore
x² = 1.8 X 10⁻⁵ X 0.35 = 0.63 X 10⁻⁵
x = 2.51 X 10⁻³ M
[OH⁻] = 2.51 X 10⁻³
pOH = -log[OH⁻] = 2.6
pH = 14- 2.6 = 11.4
b) after addition of 18 mL of HCl
The moles of acid added = molarity X volume = 0.5 x 18 = 9 mmol
the moles of base present = molarity X volume = 0.35 X 45 = 15.75 mmol
so moles of base left = 6.75 mmol
moles of salt formed = 9 mmol
This will results in formation of buffer
the pH of buffer is calculated using
pOH = pKb + log [salt] / [base]
pOH = 4.74 + log [9/6.75] = 4.86
pH = 9.14
c) on addition of 33 mL of HCl
The moles of acid added = molarity X volume = 0.5 x 33 = 16.5 mmol
the moles of base present = molarity X volume = 0.35 X 45 = 15.75 mmol
Hence the acid will completely neutralize the base
The acid left = 16.5 - 15.75 = 0.75 mmol
total volume = 33 + 45 = 75mL
[HCl] = 0.75 / 75 = 0.01 M
pH = -log[0.01] = 2
