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QUESTION

# 300 mL of 0.50 M HClO + 400 mL of 0.50m NaClO. How do I setup an ICE table for this? What is its net ionic reaction?

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

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