Question

the dissociation constant of 0.01M CH3COOH is 1.8*10^-5,then calculate the CH3COO- concentration in 0.1M HCl solution. ​

Answer 1

Answer:

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Answer 2

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