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#.

Henry Antrim · Accepted Answer

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# It is still necessary to figure out how many moles each component is. The formula for "Moles of HClO" is 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 equation for Henderson-Hasselbalch 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#

Caleb Adams · Answer

undefined The net ionic equation for the reaction between "HOCl" and "OCl"^- is:

HOCl + OCl^- -> H^+ + ClO^-

The pH of the buffer made by mixing 300 mL of 0.50 mol/L "HClO" and 400 mL of 0.50 mol/L "NaClO" can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A^-]/[HA])

Where:
pKa is the acid dissociation constant of "HClO" (hypochlorous acid)
[A^-] is the concentration of the conjugate base ("ClO^-")
[HA] is the concentration of the acid ("HClO")

Since "NaClO" is a salt of the weak acid "HClO" and its conjugate base "ClO^-", the concentrations of both "HClO" and "ClO^-" will be the same after mixing. Therefore, we can use the initial concentrations of "HClO" and "NaClO" to calculate the pH of the buffer.