# 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)#