Question

A buffer solution was prepared by mixing 391 mL of 0.431 M NaOCl and 299 mL of 0.391 M HOCl. Calculate the pH of the solution given that Ka (HOCl) is 3.2 x 10-8. Answer in 3 decimal places.?

Answer 1

The pH of the solution is 7.654.

Explanation:

Step 1. Calculate the moles of each component

`"Moles of NaOCl" = 0.391color(red)(cancel(color(black)("L NaOCl"))) × "0.431 mol NaOCl"/(1 color(red)(cancel(color(black)("L NaOCl")))) = "0.1685 mol NaOCl"`

`"Moles of HOCl" = 0.299color(red)(cancel(color(black)("L HOCl"))) × "0.391 mol HOCl"/(1 color(red)(cancel(color(black)("L HOCl")))) = "0.1169 mol HOCl"`

Step 2. Calculate the pH of the buffer

We have a solution that contains `"0.1169 mol HOCl"` and `"0.1685 mol NaOCl"`.

The equilibrium involved is

`"HOCl + H"_2"O" ⇌ "H"_3"O"^"+" + "OCl"^"-"`

A solution of a weak acid and its conjugate base is a buffer.

We can apply the Henderson-Hasselbalch equation:

`color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a₂) + log((["OCl"^"-"])/(["HOCl"]))color(white)(a/a)|)))" "`

For hypochlorous acid,

`K_text(a) = 3.2 × 10^"-8"`

`"pH" = "-log"(3.2 × 10^"-8") + log((0.1685 color(red)(cancel(color(black)("mol"))))/(0.1169 color(red)(cancel(color(black)("mol"))))) = 7.495 + log(1.44) = "7.495 + 0.159 = 7.654"`

Note I have calculated the answer to three decimal places, but I can justify only two.