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?
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?
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