# 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 $\text{HClO, p} {K}_{\textrm{a}} = 7.43$.

Apr 12, 2015

There is no net ionic reaction between $\text{HClO}$ and ${\text{ClO}}^{-}$.

#### Explanation:

This $\textcolor{red}{\text{is}}$ your buffer solution.

The buffer components are the weak acid $\text{HClO}$ and its conjugate base ${\text{ClO}}^{-}$.

The only net ionic reaction that you need is the equation for the ionization of $\text{HClO}$:

${\text{HClO" + "H"_ 2"O" ⇌ "H"_ 3"O"^+ + "ClO}}^{-}$; $\text{p"K_"a} = 7.53$

You still have to calculate the moles of each component.

$\text{Moles of HClO" = 0.300 cancel("L") × "0.50 mol"/(1 cancel("L")) = "0.15 mol}$

$\text{Moles of ClO"^(-) = 0.400 cancel("L") × "0.50 mol"/(1 cancel("L")) = "0.20 mol}$

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