Question

Ka for hypochlorous acid HCLO is 3.0x10-8. Calculate the pH after 10.0, 20.0, 30.0, and 40.0 ml of 100 m NaOH have been added to 40.0 ml of 0.100 m HCLO?

Answer 1

pH = 1/2(pKa - log C)

pH = 1/2 (-log(3.8*10^-8)-log(40*0.1/1000))

pH = 4.91

after adding NaOH then it become salt of weak acid and strong base

adding 10 mL of NaOH

pH = 7 + 1/2 (pKa - log C)

pH = 7 + 1/2 (-log(3*10^-8) - log ((100 *10/1000) - (0.1*40/1000))

pH = 10.76

adding 20 mL of NaOH

pH = 7 + 1/2 (pKa - log C)

pH = 7 + 1/2 (-log(3*10^-8) - log ((100 *20/1000) - (0.1*40/1000))

pH = 10.61

adding 30 mL of NaOH

pH = 7 + 1/2 (pKa - log C)

pH = 7 + 1/2 (-log(3*10^-8) - log ((100 *30/1000) - (0.1*40/1000))

pH = 10.523

adding 40 mL of NaOH

pH = 7 + 1/2 (pKa - log C)

pH = 7 + 1/2 (-log(3*10^-8) - log ((100 *40/1000) - (0.1*40/1000))

pH = 10.46