How many mmoles of HCl must be added to 130.0 mL of a 0.40 M solution of methylamine (pKb = 3.36) to give a buffer having a pH of 11.59?

No. of moles of CH₃NH₂ in the solution = (0.40 mmol/mL) × (130 mL) = 52 mmol

Let y mmol of HCl is added which will give y mmol of H⁺ ions (limiting reactants).
CH₃NH₂(aq) + H⁺(aq) → CH₃NH₃+(aq)

No. of moles of CH₃NH₃+ formed = y mmol
No. of moles of CH₃NH₂ unreacted = (52 - y) mmol
Hence, in the final solution, [CH₃NH₂]/[CH₃NH₃+] = (52 - y)/y

Consider the dissociation of CH₃NH₂ :
CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃+(aq) + OH⁻(aq) …. pKb

pH = pKb - log([CH₃NH₂]/[CH₃NH₃+])
14 - 11.59 = 3.36 - log([CH₃NH₂]/[CH₃NH₃+])
log([CH₃NH₂]/[CH₃NH₃+]) = 0.95
[CH₃NH₂]/[CH₃NH₃+] = 10⁰·⁹⁵
(52 - y)/y = 10⁰·⁹⁵
52 - y = 10⁰·⁹⁵y
y = 52/(1 + 10⁰·⁹⁵)
y = 5.25

Amount of HCl added = 5.25 mmol

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