Calculate the change in pH that occurs. Is this right?

Question

Just need a good explanation to see where I am going wrong.

0.05 M CH3COOH

So, CH3COOH + H2O --><-- CH3COO- + H3O+

Ka = [CH3COO-][H3O+] / [CH3COOH]

But isn't [H3O+] in this case (10-5) assumed to be 0 for these calculations? Is there a way to do this without using [H3O+] = sqrt(Ka*conc.Ha) ?

I was trying to figure out if I could calculate it using the Ka formula above.

Answer 1

You have to Use H3O = sqrt(.05*Ka)

H30 is zero initially, at equilibrium it will have a value (x)

Ka = [H30+][C2H3O2-]/[HC2H302]
Ka = [x][x]/[HC2H302]
(Ka)(HC2H302) = x^2
(Ka)(.05)^(1/2) = x = [H3O+]

Answer 2

The [H3O+] is small, but definetely not zero. Acetic acid is weak. I believe that only 4% of the acid ionizes. Don't forget that the CH3COO- will be equal to the H3O+. So your numerator will be very small and the Ka very small.

Everything you show is correct as written. Why do you sow a square root? Just use the algebra to solve for the H3o+