physical chem question!! (1 Viewer)

[thejesster]

help!! i am preparing for my chem exam next week and found a question which i can't make head nor tail of.....

a 0.100mol/L NaOH solution is used to titrate 50mL of 0.100mol/L CH3COOH. Calculate the pH of the solution after 40mL of NaOH is added [Ka = 1.8 x 10^(-5)]

please, if anyone knows, tell me how to do it!...
thank you!..

 

[Ryuu-jin-jakka]

is the pH = 12 ??? I always seems to get the wrong answer, so let me no if it is right so i can paste the solution up

 

[Ryuu-jin-jakka]

[Ka = 1.8 x 10^(-5)]

yeh just read the question just then and wtf is Ka??? O___o

 

[Ryuu-jin-jakka]

[Ka = 1.8 x 10^(-5)]

yeh just read the question fully, wtf is Ka??? O___o

 

[Trebla]

K_a is the acid dissociation constant (dissociation constants are covered in Industrial Chemistry option)

If an acid of the form HA dissociates in equilibrium as:

HA <--> H⁺ + A^-

then K_a = [H⁺] [A^-] / [HA]
where [...] means concentration

I'm assuming that the given K_a is of the acetic acid.
You know that in equilibrium:
CH₃COOH <--> CH₃COO^- + H⁺

So K_a = [H⁺] [CH₃COO^-] / [CH₃COOH]

You know that [CH₃COOH] = 0.1 M and K_a = 1.8 x 10^-5

This means that:
[H⁺] [CH₃COO^-] = 1.8 x 10^-5 x 0.1
= 1.8 x 10^-6

Since the number of moles of H⁺ and CH₃COO^- are equal (due to their stoichiometric ratio) for the same volume, then their concentrations must be equal, hence
[H⁺]² = 1.8 x 10^-6

Now all you do is take the square root to find [H⁺] and this is the concentration of H⁺ which reacts with the NaOH in the titration. Calculate the [H⁺] after the neutralisation, and then this should help calculate pH.

Not sure if this is fully correct because it has assumed the equilibrium is undisturbed throughout the reaction i.e. H⁺ concentration before neturalisation does not change, which of course is not true.

 

[Ryuu-jin-jakka]

oh if the question is related to industry chem hehe no idea then, i thought it was an acidic eninvoment question k