Answer to Question #123872 in Physical Chemistry for Claudiu

Question #123872

A buffered solution of 25 mL of an acid-base indicator is prepared by mixing 5mL of a stock solution of indicator 2,10x10^-4 M + 2,5 mL of Na2CO3 0,1 M + 10mL of NaHCO3 0,2 M
The stoichiometric equilibrium constant of carbonic acid are 4,47x10^-7 and 4,68x10^-11
Calculate the pH of this solution.

Expert's answer

The acid-base equilibrium: HCO₃^- = H⁺ + CO₃^2-

Constant is 4.68*10^-11 = [H⁺]*[CO₃^2-]/[HCO₃^-]

Concentrations in resulting solution are:

[CO₃^2-] = 0.1*2.5/(5+2.5+10) = ‭0.014 M

[HCO₃^-] = 0.2*10/(+2.5+10) = ‭0.114 M

Therefore: [H⁺] = 4.68*10^-11 * 0.114 / 0.014 = 3.8*10^-10 M

pH = -log[H⁺] = 9.42

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Assignment Expert

15.07.20, 13:56

Dear Claudiu, Questions in this section are answered for free. We can't fulfill them all and there is no guarantee of answering certain question but we are doing our best. And if answer is published it means it was attentively checked by experts. You can try it yourself by publishing your question. Although if you have serious assignment that requires large amount of work and hence cannot be done for free you can submit it as assignment and our experts will surely assist you.

Claudiu

01.07.20, 17:22

Can you also calculate the ionic strenght ?

Answer to Question #123872 in Physical Chemistry for Claudiu

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