Question

What is the net ionic equation for the reaction between `"HOCl"` and `"OCl"^"-"`? What is the pH of a buffer made by mixing 300 mL of 0.50 mol/L `"HClO"` and 400 mL of 0.50 mol/L `"NaClO"`?

For `"HClO, p"K_text(a) = 7.43`.

Answer 1

There is no net ionic reaction between `"HClO"` and `"ClO"^-`.

Explanation:

This `color(red)("is")` your buffer solution.

The buffer components are the weak acid `"HClO"` and its conjugate base `"ClO"^-`.

The only net ionic reaction that you need is the equation for the ionization of `"HClO"`:

`"HClO" + "H"_ 2"O" ⇌ "H"_ 3"O"^+ + "ClO"^(-)`; `"p"K_"a" = 7.53`

You still have to calculate the moles of each component.

`"Moles of HClO" = 0.300 cancel("L") × "0.50 mol"/(1 cancel("L")) = "0.15 mol"`

`"Moles of ClO"^(-) = 0.400 cancel("L") × "0.50 mol"/(1 cancel("L")) = "0.20 mol"`

The Henderson-Hasselbalch Equation is;

`"pH" = "p"K_"a" + log(("[ClO"^(-)"]")/("[HClO]"))`

`"pH" = 7.53 + log((0.20 cancel("mol"))/(0.15 cancel("mol"))) = 7.53 +0.12 = 7.67`