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"#?

Question

For #"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"# 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#