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
Aug 7, 2017

Answer:

#sf(pH=3)#

Explanation:

For a weak acid you can use the expression:

#sf(pH=1/2[pK_a-log[acid])#

#:.##sf(pH=1/2[6-log(1)]#

#sf(pH=3)#

Aug 7, 2017

#"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 #0.01 - "1.0 M"#), if #K_a < 10^(-5)# or so, we can consider the small #x# approximation without much loss of accuracy.

(Furthermore, the percent dissociation decreases with increasing starting concentration, so a #"1.0 M"# concentration is even better for the approximation than, say, #"0.01 M"#.)

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 #"pH"# would be:

#color(blue)("pH") = -log["H"^(+)]#

#= -log(10^(-3)) = color(blue)(3)#