Question

The pH of a solution containing 49.0 mL of 0.10 M KOH and 50.0 mL of 0.10 M of HBr is:

a. 5.0 b. 3.0 d. 7.0 e. 4.0

Answer 1

answer : b ) 3.0

solution:

millimoles of KOH = 49.0 x 0.10 = 4.90

millimoles of HBr = 50 x 0.1 = 5

millimoles of acid > millimoles of base

after reaction millimoles of acid remains = 5- 4.9 = 0.1

molarity of [H+] = 0.1 / total volume

                          = 0.1 / (49 +50)

                        = 1.0 x 10^-3 M

pH = -log [H+]

pH = -log (1.0 x 10^-3)

pH = 3.0