A 50.0-ml volume of 0.15 m hbr is titrated with 0.25 m koh. calculate the ph after the addition of 20.0 ml of koh.

Given:

Volume of HBr = 50 ml

Concentration of HBr = 0.15 m

Concentration of KOH = 0.25 m

Volume of KOH = 20 ml

Balanced Chemical Equation:

HBr + KOH = = = > KBr + H2O

pH = log ([acid]/[base]) + pKa

Ka (HBr) = 1.0 x 10^ 9

pH = log ([0.15*0.05L]/[0.25*0.020]) - log (1x10^9)

pH = 9.176