What is the pH of a solution prepared by mixing 500.mL of 0.10 M NaOCl and 500. mL of 0.20 M HOCl? [K_a(HOCl) = 3.2 × 10^-8] A) 4.10 B) 7.00 C) 7.19 D) 7.49 E) 7.80

Question

Quizplus · Accepted Answer

HOCl is a weak acid that can dissociate to form H+ and OCl-. When NaOCl and HOCl are mixed, they react to form Na+ and OCl-, which can then react with H+ to form HOCl.

The balanced chemical equation for this reaction is:

NaOCl + HOCl → Na+ + OCl- + H+

The initial concentrations of NaOCl and HOCl are 0.10 M and 0.20 M, respectively. When these solutions are mixed, the total volume of the solution becomes 500 mL + 500 mL = 1000 mL = 1 L.

To calculate the pH of the solution, we first need to find the concentrations of OCl- and HOCl at equilibrium. Since the reaction is a weak acid equilibrium, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([OCl-]/[HOCl])

where pKa is the acid dissociation constant, [OCl-] and [HOCl] are the equilibrium concentrations of OCl- and HOCl, respectively.

The pKa value for HOCl is given as 3.2 × 10-8. We can find the equilibrium concentrations of OCl- and HOCl using the following equations:

[OCl-] = moles of NaOCl reacted / total volume of solution

[HOCl] = (initial [HOCl] - moles of NaOCl reacted) / total volume of solution

The moles of NaOCl reacted can be found using the balanced chemical equation:

1 mole of NaOCl reacts with 1 mole of HOCl to form 1 mole of OCl- and 1 mole of H+

Since the initial volumes of NaOCl and HOCl are equal, the moles of NaOCl reacted is equal to half the initial moles of NaOCl:

moles of NaOCl reacted = (0.10 M) x (500 mL / 1000 mL) x (1 L / 1000 mL) x (1/2) = 0.025 moles

Using this value, we can now calculate the equilibrium concentrations of OCl- and HOCl:

[OCl-] = moles of NaOCl reacted / total volume of solution = 0.025 moles / 1 L = 0.025 M

[HOCl] = (initial [HOCl] - moles of NaOCl reacted) / total volume of solution

= (0.20 M - 0.025 moles / 1 L) = 0.175 M

Now we can use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log([OCl-]/[HOCl])

= 3.2 × 10-8 + log(0.025 M / 0.175 M)

= 7.19

Therefore, the pH of the solution is 7.19, which is answer choice C.

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