Question

Calculate the pH during the titration of 25.00 mL of 0.1000 M HBrO(aq) with 0.1000 M LiOH(aq) after 16 mL of the base have been added. Ka of hypobromous acid = 2.3 x 10-9.

Answer 1

Given:

M(HBrO) = 0.1 M

V(HBrO) = 25 mL

M(LiOH) = 0.1 M

V(LiOH) = 16 mL

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

mol(HBrO) = 0.1 M * 25 mL = 2.5 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.1 M * 16 mL = 1.6 mmol

We have:

mol(HBrO) = 2.5 mmol

mol(LiOH) = 1.6 mmol

1.6 mmol of both will react

excess HBrO remaining = 0.9 mmol

Volume of Solution = 25 + 16 = 41 mL

[HBrO] = 0.9 mmol/41 mL = 0.022M

[BrO-] = 1.6/41 = 0.039M

They form acidic buffer

acid is HBrO

conjugate base is BrO-

Ka = 2.3*10^-9

pKa = - log (Ka)

= - log(2.3*10^-9)

= 8.638

use:

pH = pKa + log {[conjugate base]/[acid]}

= 8.638+ log {3.902*10^-2/2.195*10^-2}

= 8.888

Answer: 8.89