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18 Dec 2019

For hypochlorous acid, HOCl, Ka= 3.5 X 10-4. What [OCl-]/ [HOCl] ratio results in a buffer with a pH 7.70?

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Deanna Hettinger
Deanna HettingerLv2
31 Dec 2019

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Related questions

Hypochlorous acid, HOCl, reacts with water as shown below:

HOCl(aq) + H2O(l) <---> OCl-(aq) + H3O+(aq)

What is the pH of a 0.033 M aqeous solution of acetic acid? Ka=1.1e-2

Could you please help me to solve (iii)

You are disinfecting drinking water with chlorine to give a final concentration in the water of 1mg.l-1 as free chlorine. There is no chlorine demand. The pH of the water is 7.87 and the pKa for hypochlorous acid is 7.53. The reaction is:

HOCl = H + OCl

The version of the Henderson-Hasselbalch equation is:

pH= pKa - log ([HA]/[A])

(i) Calculate the ratio of Hypochlorous acid to the hypochlorite anion (HOCl/OCl- ).

My answer: pH= pKa - log ([HA]/[A])

[HOCl]/[OCl]=107.53-7.87pH

[HOCl]/[OCl]=10-0.34=0.457

(ii) Calculate the percentage as HOCl from the above results

My answer: 0.457/(0.457+1) *100= 31.4% of HOCl

(iii) Using the same conditions, but with a chlorine demand for this water of 0.2mg.l-1 of free chlorine, what would be the free chlorine residual?

What is the ph of a 0.10 M solution of hypochlorous acid, HOCl?Ā  Ka of HOCl is 3.5Ɨ10āˆ’8 at 250C.

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