a. calculate the ph of a buffer that is prepared by mixing 25.0 ml of 0.300 m methylamine (ch3nh2) and 0.405 g of methylammonium chloride with water to make 500.0 ml of solution.

pH = 10.74 (The following uses two methods; common ion effect & Henderson-Hasselbalch Equation for weak base buffers)

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