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

Question

For $HClO, p K_{a} = 7.43$ $"HClO, p"K_text(a) = 7.43$ .

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Answer 1 · 2015-04-12T20:25:59

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

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

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

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

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

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;

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

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