Question

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?

Answer 1

`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)`

Answer 2

`"pH" = 3`.

If you wish, you could use the equation shown at the end here, which is what Michael has done.

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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)`