given 0.5 for CH3COOH initial concentration, and pH of 2.38 , what is the Ka , H+ and pKa?

Answer 1

I presume your 0.5 is moles/liter; please be more careful next time. The equilibrium reaction is
HOAc -> OAc- + H+
Some HOAc dissociates to form enough H+ to bring the pH from 7 (neutral water) to 2.38. Call this moles/liter = x. OAc- also is x. Generally, x will be much smaller than HOAc, so HOAc- x
appx. = HOAc. We have
x^2/ 0.5 = Ka
We use the pH to find [H+].
[H+] = 10^-pH = 10-3 + 10^0.62 = 4.1x10-3 (appx)
Then x^2/0.5 = 17 x 10-6/0.5 = 3.4x10^-5 = Ka
To find pKa, we use the same definition as for pH,
pKa = - Log (Ka) = 5 - Log 3.4 = 4.45 (appx)

Answer 2

pH = 2.38
[H+] = 10^-2.38 = 4.17*10^-3
Acetic Acid dissociates into both H+ and C2H3O2(-) equally, therefore the concentration of C2H3O2 is the same as the H+ concentration.
[C2H3O2-] = 4.17*10^-3
Now, the Ka = ([H+]*[C2H3O2])/([CH3COOH]) which you can write as Ka = {[x][x])/([.5 M - x]) where x = 4.17*10^-3
Therefore, Ka = (4.17*10^-3)^2/(.5 - 4.17*10^-3) = 3.5*10^-5
the pKa is merely the negative log of that which is 4.45

The Ka is slightly different than the listed value for acetic acid (1.8*10^-5) but this can be possibly be accounted for by any differences in temperature since the listed value is for 298 K.

Answer 3

Assuming that at equilibrium the dissociation of H+ = C6H5COO- Ka = [H+][C6H5COO-]/[C6H5COOH] = [H+]^2/[C6H5COOH] so [C6H5COOH] = [H+]^2/Ka concentration of susceptible acid, C6H5COOH = ( (10^-2.fifty 4)^2) / 6.3x10^-5 = 0.132 moldm^-3