The Kb of dimethylamine is 5.9 x 10^-4 at 25 C. Calculate the pH of a 1.20 x 10^

Question

The Kb of dimethylamine is 5.9 x 10^-4 at 25 C. Calculate the pH of a 1.20 x 10^-3 M solution of dimethylamine. In this problem, x is not small compared to the concentration and it can't be ignored.

The correct answer is 10.77, and not 10.93. How would I do this since "x" can't be ignored? Quadratic equation? Every time I try I get imaginary numbers. I am not doing something right. Help? The Kb of dimethylamine is 5.9 x 10^-4 at 25 C. Calculate the pH of a 1.20 x 10^-3 M solution of dimethylamine. In this problem, x is not small compared to the concentration and it can't be ignored.

The correct answer is 10.77, and not 10.93. How would I do this since "x" can't be ignored? Quadratic equation? Every time I try I get imaginary numbers. I am not doing something right. Help?

The correct answer is 10.77, and not 10.93. How would I do this since "x" can't be ignored? Quadratic equation? Every time I try I get imaginary numbers. I am not doing something right. Help?

Explanation / Answer

Kb = [HB+][OH-]/[B]

Kb =5.9*10^-4

[HB+] = x = [OH-]

[B] = 1.2*10^-3 - x ( we must account, since problem statement)

then

Kb = [HB+][OH-]/[B]

5.9*10^-4 = x*x/(1.2*10^-3 - x)

NOTE: quadratic expression gets as:

(1)x^2 + (5.9*10^-4)x + (-1.2*10^-3)(5.9*10^-4 ) = 0

personal tip:

x^2 + Kb(x) - M*Kb = 0

be sure to consider the negative, then you will get imaginary

x = OH = 5.96*10^-4

pOH = -log(5.96*10^-4 = 3.224753

pH = 14-3.224753= 10.775

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