The Ka value of a weak acid, HA, is 10-6. An initial solution containing 1.0 M of HA is prepared. What is the pH of the final solution at equilibrium?
2 Answers
Explanation:
For a weak acid you can use the expression:
#"pH" = 3# .
If you wish, you could use the equation shown at the end here, which is what Michael has done.
For this weak acid, a general equilibrium can be written:
#"HA"(aq) rightleftharpoons "H"^(+)(aq) + "A"^(-)(aq)#
After you make your ICE table, or whatever floats your boat, you should end up with a mass action expression of:
#K_a = 10^(-6) = (x^2)/(1.0 - x)#
For typical concentrations (about
(Furthermore, the percent dissociation decreases with increasing starting concentration, so a
This says that we can write:
#K_a ~~ (x^2)/(1.0)#
or that
#x -= ["H"^(+)] = sqrt(K_a("1.0 M"))#
#= sqrt(10^(-6) cdot 1)#
#= 10^(-3)# #"M"#
And thus, the
#color(blue)("pH") = -log["H"^(+)]#
#= -log(10^(-3)) = color(blue)(3)#