If the Ka of a monoprotic weak acid is 4.5x10^-6 what is the pH of a .10M solution of this acid.

I don't even know how to start this problem, I'm totally lost please help!

Step 1. Write and balance the equation.

Step 2. Complete the ICE chart (initial, change. equilibrium).
Step 3. Write the Ka expression.
Step 4. Substitute the E line of the ICE chart into Ka expression and solve for the unknown.

.........HA ==> H^+ + A^-
I........0.1....0......0
C........-x.....x......x
E......0.1-x....x......x

Ka = (H^+)(A^-)/(HA)

4.5E-6 = (x)(x)/(0.1-x)
Solve for x = (H^+), then convert to pH.

ph=10.3

The pH is not 10.3

I got that the pH equals 2.67 is this number correct?

I don't get 2.67. If you will show your work I will find the error. I agree the answer isn't 10.3.

The correct pH is 3.17

3.17 is correct.