How to Calculate the pH After Adding NaOH to CH3COOH?

[dumbadum]

The question is:
calculate the pH after 10.0 mL of 0.4 M NaOH is added to 20.0 mL of 0.5 M CH3COOH (Ka CH3COOH = 1.8 x 10^-5)

I'm thinking that if I find the Kb of NaOH, then I can find how much of OH and H will be produced. I can then subtract the concentrations and find the pH or something. But how do I take into account the fact that CH3COONa is a basic salt and how to I find the Kb or NaOH?
Or in short, How would I go about the question?
Thank you!

 

[ksinclair13]

Well, the reaction that will take place will be (I wrote the inorganic formula for acetic acid):

HC₂H₃O₂ + OH^- ---> C₂H₃O₂^- + H₂O

Now, figure out how many millimoles of each reactant and how many milliliters of each reactant. Then, do the stoichiometry to figure out how many mmol and mL there are of the conjugate base and acid after the reaction goes to completion. Then, calculate the molarity of the conjugate base and the acid, and then you should be able to use the Henderson-Hasselbach equation to figure out the pH.

I hope this works =)

 

[siricon]

goes like that

i guess it goes like that...
you have a solution 0,5 M of CH3COOH
this solution has 20 mL meaning u have 0,01 mol of the acid... when u mix it with the 10 mL from the NaOH the new solution will have a volume of 30 mL... so the new molarity of the acid is 0,33 (0,01 mol/0,03 L)
doing the same with the NaOH you find out u have 4.10^-3 mol, meaning the new solution will have a molarity of 0,13
the acid concentration is higher than the base´s; since they are together in the same recipient there will happen a neutralization reaction between them; since there is more acid than base, the final concentration of acid will be 0,01 - 4.10^-3 = 6.10^-3 moles of the acid in 30 mL..the final molarity will be 0,2
then u calculate it normaly
Ka = ([H+[ . [CH3COO-[) / [CH3COOH[
1,8.10^-5 . 0,2 = X^2
x = 1,897.10^-3
pH = 2,72

 

[pizzasky]

(1) K_{b} is an expression used only for weak bases. Since NaOH is a strong base (as it dissociates completely in water), it has no K_{b} value.

(2) You should know that [H^+] = \frac{K_{a}[CH_{3}COOH]}{[CH_{3}COO^-]}. So, can you figure out [CH_{3}COO^-]\ and\ [CH_{3}COOH] once the neutralisation reaction is complete?

(3) You are right in saying that CH_{3}COONa is a basic salt. However, CH_{3}COO^-\ and\ CH_{3}COOH exist in equilibrium, and the K_{a} formula has already taken into account the basicity of CH_{3}COO^- when calculating the pH, so one formula is all you need!

HINT: NaOH is the limiting reagent in the neutralisation reaction.

NOTE: When calculating the final concentrations, remember that the total volume of the solution has changed.

POSTSCRIPT: (Further explanation for comment 3) Observe the K_{a} formula above. Note that if [CH_{3}COO^-] increases, [H^+] will decease and the pH will increase. The reverse is also true. This is why I say that "the K_{a} formula has already taken into account the basicity of CH_{3}COO^-"

Hope this helps!