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
Explanation:
#"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
Initially, before dissociation occurs, the concentration of each species, in
#["HA"] = 0.01# #["H"^+] = 0# #["A"^-] = 0#
After
#["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
#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
#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
#["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#