The p𝐾a is 7.530 . A 53.0 mL solution of 0.132 M NaOCl is titrated with 0.271 M HCl . Calculate the pH of the solution after the addition of 7.77 mL of 0.271 M HCl .

Question

The p𝐾a of hypochlorous acid is 7.530. A 53.0 mL solution of 0.132 M sodium hypochlorite (NaOCl) is titrated with 0.271 M HCl. 

Calculate the pH of the solution after the addition of 7.77 mL of 0.271 M HCl.

Calculate the pH of the solution after the addition of 27.2 mL of 0.271 M HCl.

Calculate the pH of the solution at the equivalence point with 0.271 M HCl.

Answer 1

I will do the first calculation. The others are done the same way, just changing the numbers. In the ICE table below, I have used mmoles instead of moles (e.g. 6.99 mmols ClO- instead of 0.00699 mols) just to save space. In the final row of the table, I've converted back to molar concentrations. Hopefully you'll be able to follow what I've done. If not, leave a comment and hopefully I'll be able to respond. (BTW, you really only need 3 sig. figs)

ClO- + H+ ==> HClO

6.99.....2.11.........0.........I

-2.11....-2.11.....+2.11....C

4.88........0..........2.11....E ... volume = 53 + 7.77 ml = 60.77 ml = 0.06077 L

0.0803...............0.0347 M ... concentrations in M

pH = pKa + log [ClO-] / [HClO]

pH = 7.530 + log (0.0803/0.0347) = 7.530 + 0.3644

pH = 7.894

At the equivalence point, pH = pKa so that's a simple one.