How many grams of solid potassium hypochlorite should be added to 2.00 L of a 0.244 M hypochlorous acid solution to prepare a buffer with a pH of 6.733? ka = 3.5*10-8

grams potassium hypochlorite = g.

8.2763 g

Explanation:

Considering the Henderson - Hasselbalch equation for the calculation of the pH of the buffer solution as:

pH=pKa+log[base]/[acid]

Where Ka is the dissociation constant of the acid.

Given that the acid dissociation constant = 3.5*10⁻⁸

pKa = - log (Ka) = - log (3.5*10⁻⁸) = 7.46

Given concentration of acid = [acid] = 0.244 M

pH = 6.733

So,

6.733 = 7.46+log[base]/0.244

[Base] = 0.0457 M

Given that Volume = 2 L

So, Moles = Molarity * Volume

Moles = 0.0457 * 2 = 0.0914 moles

Molar mass of potassium hypochlorite = 90.55 g/mol

Mass = Moles * Molar mass = (0.0914 * 90.55) g = 8.2763 g