Question #58e5e

1 Answer
Jun 8, 2017

Answer:

#3.5xx10^-8#

Explanation:

We're asked to find the acid-dissociation constant #K_a# of hypochlorous acid, #"HOCl"# with a given molar concentration and #"pH"#.

Let's first write the equilibrium reaction equation:

#"HOCl" (aq) rightleftharpoons "H"^+ (aq) + "ClO"^(-) (aq)#

From this, we can write the equilibrium constant expression:

#K_a = (["H"^+]["ClO"^(-)])/(["HOCl"]#

From the given #"pH"# we can calculate the hydrogen-ion concentration, #["H"^+]#:

#["H"^+] = 10^(-"pH") = 10^(-3.9) = 1.26 xx 10^-4 M#

Let's create a makeshift I.C.E. chart using bullet points to illustrate the equilibrium changes:

Initial

  • #"HOCl"#: #0.45M#

  • #"H"^+#: #0M#

  • #"ClO"#: #0M#

Change

  • #"HOCl"#: #-1.26xx10^-4M#

  • #"H"^+#: #+1.26 xx 10^-4M#

  • #"ClO"#: #+1.26 xx 10^-4M#

Final

  • #"HOCl"#: #0.45M - 1.26xx10^-4M#

  • #"H"^+#: #1.26xx10^-4M#

  • #"ClO"#: #1.26xx10^-4M#

From these equilibrium concentrations, we can plug these in to the equilibrium expression to calculate the #K_a#:

#K_a = ([1.26xx10^-4M][1.26xx10^-4M])/([0.45M-1.26xx10^-4M]) = color(red)(3.5xx10^-8#