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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 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.

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter15: Additional Aqueous Equilibria
Section: Chapter Questions
Problem 110QRT: Consider the nanoscale-level representations for Question 110 of the titration of the aqueous weak...
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The pKₐ of hypochlorous acid is 7.530. A 51.0 mL solution of 0.122 M sodium hypochlorite (NaOCl) is titrated with 0.323 M HCl. 1. **Calculate the pH of the solution after the addition of 7.44 mL of 0.323 M HCl.** pH = 7.73 2. **Calculate the pH of the solution after the addition of 20.0 mL of 0.323 M HCl.** pH = 2.475 3. **Calculate the pH of the solution at the equivalence point with 0.323 M HCl.** pH = ___ (Note: The pH at the equivalence point is not provided and needs to be calculated based on the complete neutralization in the titration process.)
Transcribed Image Text:The pKₐ of hypochlorous acid is 7.530. A 51.0 mL solution of 0.122 M sodium hypochlorite (NaOCl) is titrated with 0.323 M HCl. 1. **Calculate the pH of the solution after the addition of 7.44 mL of 0.323 M HCl.** pH = 7.73 2. **Calculate the pH of the solution after the addition of 20.0 mL of 0.323 M HCl.** pH = 2.475 3. **Calculate the pH of the solution at the equivalence point with 0.323 M HCl.** pH = ___ (Note: The pH at the equivalence point is not provided and needs to be calculated based on the complete neutralization in the titration process.)
Expert Solution
Step 1

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

 

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