Question

What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)

5.5 x 10-5

4.0 x 10-6

6.9 x 10-4

3.5 x 10-10

4.0 x 10-11

Answer 1

Ka =
`4.04 * 10^(-11)`

Step-by-step explanation:

Initial concentration of weak acid =
`4.5 * 10^(-4)\ M`

pH = 6.87

`pH = -log[H^+]`

`[H^+]=10^(-pH)`

`[H^+]=10^(-6.87)=1.35 * 10^(-7)\ M`

HA dissociated as:

`HA \leftrightharpoons H^+ + A^(-)`

(0.00045 - x) x x

[HA] at equilibrium = (0.00045 - x) M

x =
`1.35 * 10^(-7)\ M`

`Ka = ([H^+][A^(-)])/([HA])`

`Ka = ((1.35 * 10^(-7))^2)/(0.00045 - 0.000000135)`

0.000000135 <<< 0.00045

`Therefore, Ka = ((1.35 * 10^(-7))^2)/(0.00045 ) = 4.04 * 10^(-11)`