Calculate the hydronium ion concentration of a 0.250M solution of ethaonic acid (Ka= 1.82 x 10^-5)

Answer 1

HAc = CH3COOH = ethanoic acid.
.................HAc ==> H^+ + Ac^-
I................0.250.......0........9
C..............-x..............x........x
E...........0.25-x..........x........x

Write the Ka expression for HAc and plug in the E line. Solve for x = (H^+).

Answer 2

I am a little bit confused with it please...thanks for the help...

Answer 3

My explanation looks simple to me.
Ka = 1,82E5 = (H^+)(Ac^-)/(HAc)
1.82E-5 = (x)(x)/(0.250-x)
Solve for x. What's confusing.

Answer 4

I need the , quite confused