Question

The Kb of hydroxylamine, NH2OH, is 1.10 x10-8. A buffer solution is prepared by mixing 120 mL of a 0.36 M hydroxylamine solution with 50 mL of a 0.28 M HCl solution.


Find the pH of the resulting solution.

Answer 1

Answer:

pH = 5.9

Explanation:

NH₂OH + HCl => NH₂Cl + H₂O

=> 120ml(0.36M NH₂OH) + 50ml(0.28M HCl)

=> 0.12L(0.36M) NH₂OH + 0.05L(0.28M) HCl

=> 0.0432 mole NH₂OH + 0.014 mole HCl

=> (0.0432 - 0.014) mole NH₂OH + 0.0432 mole NH₂Cl

=> 0.0292 mole NH₂OH + 0.0432 mole NH₂Cl in 170 ml of solution

=> (0.0292mol/0.17L) NH₂OH + (0.0432mol/0.17L) NH₂Cl

=> (0.172M) NH₂OH + (0.254M) NH₂Cl

=> in water 0.172M HONH₃OH + 0.254M HONH₃Cl

        HONH₃OH              ⇄          HONH₃⁺           +         OH⁻

C(i)      0.172M                               0.254M                      0.0 M

ΔC          -x                                         +x                             +x

C(eq) 0.172-x = 0.172M         0.254+x = 0.254M                x  

Kb = [HONH₃⁺][OH⁻]/[HONH₃OH] = (0.254)[OH⁻]/(0.172) = 1.1 x 10⁻⁸

=> [OH⁻] = (1.1 x 10⁻⁸)(0.172)/(0.254)M = 7.5 x 10⁻⁹M

=> [H⁺} = Kw/[OH⁻] = (1 x 10⁻¹⁴/7.5 x 10⁻⁹)M = 1.33 x 10⁻⁶M

pH = -log[H⁺] = -log(1.33 x 10⁻⁶) = 5.9

Calculation check using base form of Henderson-Hasselbalch equation:

pOH = pKb + log([acid]/[base])

pKb = -log(Kb) = -log(1.1 x 10⁻⁸) = 7.9

pOH = 7.9 + log(0.254/0.172) = 7.9 + log(1.48) = 7.9 + 0.17 = 8.13

pH + pOH = 14 => pH = 14 - pOH = 14 - 8.13 = 5.87 ≅ 5.9 (check!)