Consider a 0.400 M CH3COOH solution (Ka=1.8*10^-5) and determine the [H3O+] concentration at equilibrium

Question

Question

Answer 1

Answer #1

We know

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

Let X= [CH3COO-][H3O+]

1.8×10^-5= X * X/ (0.4-X)

X² = 1.8×10^-5 (0.4-x)

On solving this quadratic equation we know that get

X=[H3O+] = 0.0002674 = 2.674×10^-4

Answer 2

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