Answer:
pH = 10.74 (The following uses two methods; common ion effect & Henderson-Hasselbalch Equation for weak base buffers)
Step-by-step explanation:
- Buffer solution => 25 ml (0.300M CH₃NH₃OH) + 0.405 gms CH₃NH₃Cl
- fm literature Kb(CH₃NH₂) = 4.4 x 10⁻⁴ and f.wt. CH₃NH₃Cl = 67.52 g/mol.
- pKb = -log(Kb) = -log(4.4 x 10⁻⁴) = -(-3.36) = 3.36
- Concentration of CH₃NH₃Cl = (0.405g/67.52g/mol)·(0.025L)⁻¹ = 0.240M
Wk Base in aqueous media
CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻
Common Ion Effect:
CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻
C(i) 0.300M 0.240M 0
ΔC -x +x +x
C(eq) 0.300 - x 0.240 + x
≅ 0.300M* ≅ 0.240M* x => *Conc/Kb > 100 => drop 'x'
Kb = [CH₃NH₃⁺][OH⁻]/[CH₃NH₃OH ]
4.4 x 10⁻⁴ = (0.240M)[OH⁻]/(0.300M)
=> [OH⁻] = (4.4 x 10⁻⁴)(0.300M)/(0.240M) = 5.5 x 10⁻⁴M
pOH = - log[OH⁻] = -log(5.5 x 10⁻⁴M) = -(-3.26) = 3.26
pH + pOH = 14 => pH = 14 - pOH = 14 - 3.26 = 10.74
Henderson-Hasselbalch Equation for Weak Base Buffers:
pOH = pKb + log([Conj Acid]/[Wk Base])
Using above data...
pOH = pKb(CH₃NH₂) + log ([CH₃NH₃⁺]/[CH₃NH₃OH])
pOH = 3.36 + log[(0.240)/(0.300)] 3.36 + (-0.097) = 3.26
pH = 14 - pOH = 14 - 3.26 = 10.74