Calculate [ H + ] of a 0.200 M solution of Aniline C6H5NH2 Kb = 7.40e-10?

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Calculate [ H + ] of a 0.200 M solution of Aniline C6H5NH2 Kb = 7.40e-10?

2007-03-11 15:59:59 · 1 answers · asked by Teddy JJ 1 in Science & Mathematics ➔ Chemistry

1 answers

Kb = [C6H5NH3+][OH-]/[C6H5NH2]
7.4*10^-10 = x^2/0.200-x

Since 0.200 >>>7.4*10^-10, x will be insignificant compared to 0.200

7.4*10^-10 = x^2/0.200
1.48*10^-10 = x^2
x = 1.22 *10^-5 = [OH-]
[H+] = 1.0*10^-14/1.22*10^-5 = 8.22*10^-10

2007-03-11 16:08:26 · answer #1 · answered by TheOnlyBeldin 7 · 0⤊ 0⤋