Question

A sample of 0.320 M hydroxylamine (HONH2) is reacted with 0.210 M HBr solution. Kb =1.1 x 10-8 for hydroxylamine

What is the pH of the initial hydroxylamine solution?

What is the pH of the initial HBr solution?

What is the pH when you mix 25.0 mL of HONH₂ with 25.0 mL of HBr.

Answer 1

1) initially only weak base present

for weak bases

pOH = 1/2 [pKb - log C]

pKb = -log [Kb] = - log [1.1 x 10^-8]

pKb = 7.96

pOH = 1/2[7.96 - log 0.320]

pOH = 4.23

pH = 14 -4.23

pH = 9.77

2) initially HBr = 0.210 M

[H⁺] = [HBr] = 0.210 M

pH = - log [H⁺]

pH = - log [0.210]

pH = 0.68

3) millimoles of HONH2 = 25.0 x 0.320 = 8.0

millimoles of HBr added = 25.0 x 0.210 = 5.25

after HBr added

millimoles of HONH2 = 8.0 - 5.25 = 2.72

millimoles of HoNH3⁺ = 5.25

[HONH2] = 2.72 / 50 = 0.054 M

[HONH3⁺] = 5.25 / 50 = 0.105 M

pOH = pKb + log [HONH3⁺ ] / [HONH2]

pOH = 7.96 + log [0.105] / [0.054]

pOH = 8.25

pH = 14 - 8.25

pH = 5.75