The pK, of hypochlorous acid is 7.530. A 51.0 mL solution of 0.122 M sodium hypochlorite (NaOCI) is titrated with 0.323 M HCI. Calculate the pH of the solution after the addition of 7.44 mL of 0.323 M HCI. pH = 7.73 Calculate the pH of the solution after the addition of 20.0 mL of 0.323 M HCI. pH = 2.475 Calculate the pH of the solution at the equivalence point with 0.323 M HCl.
The following reaction occurs in this titration
OCl- + H3O+ HOCl + H2O ..... (1)
(hypochlorite) (hypochlorous acid)
pka of HOCl = 7.530
According to question, 51ml 0.122 M sodium hypochlorite (NaOCl) = (51ml) x (0.122mole/lt) = 6.222 mmole of NaOCl
i) after addition of 7.44 ml 0.323M HCl is added
so, no of mmoles of H3O+ = (7.44ml) x(0.323mol/lt) = 2.403mmole
According to eq (1) no of mmole of OCl- remains unreacted = (6.222-2.403)mmole = 3.819mmole.
No of mmole of HOCl formed = 2.403mmole.
Here sodium hypochlorite(NaOCl) solution is titrated with HCl to form HOCl. So we will apply Hendersons equation to find pOH of the solution.
pOH = pkb + log([HOCl]/[OCl-])
= (14 - pka) + log(2.403/3.819)
= 6.47 + (-0.201)
= 6.269
So, pH of the solution is (14-6.269) = 7.731
Step by step
Solved in 3 steps
