Sodium hypochlorite (NaOCl) is the active ingredient of bleach. At a typical working strength of 8.25%, solid

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Question:

Sodium hypochlorite (NaOCl) is the active ingredient of bleach. At a typical working strength of 8.25%, solid NaOCl completely dissociates in aqueous solutions:

NaOCl(s) ⟶ Na+(aq) + OCl-(aq)

and the following equilibrium forms between the hypochlorite (OCl-) from the dissociation of NaOCl and hypochlorous acid (HOCl):

HOCl(aq) + H2O(l) ⇆ OCl-(aq) + H3O+(aq)

2a) The pKa of HOCl is 7.53. Assume 1.24 g of solid NaOCl is dissolved in an aqueous solution buffered to pH 7.15 and gives a total volume of 1.50 L. Find the ratio [OCl-]/[HOCl]. To receive credit, you must show all work in a legible, highly organized manner.

2b) What would the ratio [OCl-]/[HOCl] be if the final volume was 3.00 L instead of 1.50 L? In complete sentences, clearly justify your reasoning.