Question

calculate the pH when 50.0ml of 0.150M KOH is mixed with 20.0mL of 300.0 M HBrO(Ka=2.5x10^-9)

Answer 1

Given:

M(HBrO) = 0.3 M

V(HBrO) = 20 mL

M(KOH) = 0.15 M

V(KOH) = 50 mL

mol(HBrO) = M(HBrO) * V(HBrO)

mol(HBrO) = 0.3 M * 20 mL = 6 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.15 M * 50 mL = 7.5 mmol

We have:

mol(HBrO) = 6 mmol

mol(KOH) = 7.5 mmol

6 mmol of both will react

excess KOH remaining = 1.5 mmol

Volume of Solution = 20 + 50 = 70 mL

[OH-] = 1.5 mmol/70 mL = 0.0214 M

use:

pOH = -log [OH-]

= -log (2.143*10^-2)

= 1.669

use:

PH = 14 - pOH

= 14 - 1.669

= 12.331

Answer: 12.33