Question

Dissociation constant of 0.01 M ch3cooh is 1.8×10^-5. Then calculate the dissociation constant of ch3coo-
1) 5.6×10^-12
2)5.6×10^-8
3) 5.6×10^-10
4)1.8×10^-9

Answer 1

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The concentration of acetate ion in a 0.01 M solution of acetic acid mixed with 0.1 m of HCl is equal to 1.8*10^-6 M.

Let a be the degree of dissociation of acetic acid and c be the initial concentration of acetic acid which is equal to 0.01 M.

So, at equilibrium the concentration of acetic acid is c(1-a), the concentration of acetate ion is ca and the concentration of proton is (0.1+ca) [ 0.1 is contributed by HCl which dissociates completely and ca is the contribution from the acetic acid].

Now, Ka = [H+].[CH3COO-]/[CH3COOH]

We know that, a<<1 and ca<<0.1, so the equation simplifies to ka = 0.1.a.

Hence, a = ka/0.1 = 1.8*10^-4

therefore, [CH3COO-] is equal to ca = 1.8*10^-6 M