The #K_"a"# for the weak acid HA is #4.0xx10^-6#. what is the pH of a #0.01"M"# solution of the acid? what is its #"p"K_"a"#?

1 Answer
Mar 30, 2016

#"p"K_"a" = 5.4#

#"pH" = 3.7#

Explanation:

#"p"K_"a"# is simply #-log_10(K_"a")#. It is a constant at any temperature and does not depend on the molarity of the solution.

#"p"K_"a" = -log_10(K_"a")#

#= -log_10(4.0xx10^{-6})#

#= 5.4#

To find the pH of the solution, we let the extent of dissociation be #x#.

Initially, before dissociation occurs, the concentration of each species, in #"M"#, are as follows.

  • #["HA"] = 0.01#
  • #["H"^+] = 0#
  • #["A"^-] = 0#

After #x# amount of dissociation has occurred, the solution is in equilibrium. The concentration of each species, in #"M"#, are as follows.

  • #["HA"] = 0.01-x#
  • #["H"^+] = x#
  • #["A"^-] = x#

Since they are in equilibrium, the relation below holds

#K_"a" = frac{["H"^+] * ["A"^-]}{["HA"]}#

#4.0xx10^{-6} = frac{x * x}{0.01-x}#

This is a quadratic equation. But note that since #K_"a"# is small, the acid is a weak acid. Therefore, dissociation occurs to a little extent only. Hence, we can assume #x "<<" 0.01# to simplify our calculations.

#4.0xx10^{-6} = frac{x^2}{0.01-x}#

#~~ frac{x^2}{0.01}#

Thus,

#x = sqrt{(4.0xx10^{-6}) xx 0.01}#

#= 2.0 xx 10^{-4}#

Our original assumption of #x "<<" 0.01# is valid and we can check that

#frac{(2.0 xx 10^{-4})^2}{0.01-(2.0 xx 10^{-4})} = 4.008 xx 10^{-6}#

#~~ 4.0 xx 10^{-6}#

Since we know #x#, we know that at equilibrium,

#["H"^+] = x = 2.0 xx 10^{-4}#

The pH of the solution is given by

#"pH" = -log_10(["H"^+])#

#= -log_10(2.0 xx 10^{-4})#

#= 3.7#