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 CH₃NH₃Clfm 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) = 3.36Concentration 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

                 

Each names a given number of objects

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