Calculate the pH of a 0.0396 M solution of a weak acid whose ionization constant Ka is 9.54 x 10^-3

Every time I solve for [H+], I end up with a negative number, and since you can't take the negative log of a negative number to find pH, I can't solve this!

Did I do something wrong to end up with a negative number for [H+]?
Thanks in advance!

answering your email.....

HA <= = = > H+ & A-
(0.0396 -X) <=> X & X

an acid with a Ka the size of "10^-3" ionizes significantly,
to the point that we can not simplify the concentration of the acid,
from (0.0396 -X) to just (0.0396)

Ka = [H+] [A-] / [HA]

9.54 X 10^-3 = [X] [X] / [0.0396 -X]

9.54 X 10^-3 [0.0396 -X] = X2

0.0003778 - 9.54 X 10^-3 [X] = X2

X2 + 9.54 X 10^-3 [X] - 0.0003778 = 0

using the quadratic equation at
http://www.math.com/students/calculators…
X = [H+] = 0.0152 Molar

pH = - (log of 0.0152 Molar)

pH = - (-1.818)

pH = 1.818

I don't know what web assign wants you to do adout rounding off
your data came with 3 sig figs in "0/0396M) & 3 sig figs in "9.64"
a pH does not begin to have sig figs until after the decimal point
the pH of "1.818" has 3 sig figs

http://www.ndt-ed.org/GeneralResources/S…

so remember Ka = Ka = ([H+]*[A-])/([HA])

remember [H+] = [A-]

so we can rewrite it as:

Ka = ([H+]^2)/([HA])

so we can further rearrange it to:

[H+] = sqrt( [HA] * Ka)

remember HA is concentration of acid

[H+] = sqrt( 3.96e-2 * 9.54e-3)

[H+] = 1.9e-2

pH = -log (1.9e-2)

pH = 1.7

edit: fixed a typo