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
