A student prepared a 0.10 M solution of acetic acid, CH_3COOH. Acetic acid has a K_a of 1.75 x 10^-3.?

what is the hydronium ion concentration of the solution ?

Answer:
CH3COOH + H2O <----- > CH3COO- + H3O+
initial concentration
0.10
change
- x . .. . . . . .. . . . . . .. . . . . . . .. . .+x .. . . . . . .+x
at equilibrium
0.10-x . .. . . . . . . . . . . .. . . . . . . . .x . .. . . . . . .x

1.75 x 10^-3 = (x)(x) / 0.1 - x
x = [H3O+] = 0.0132 M

I know that Ka of acetic acid is 1.75 x 10^-5 ( at 25 °C) .
in this case [H3O+] = 0.00132 M

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