pH calculation I struggle with

[Nigthwishxd]

Hi! I have tried to solve this for few days but can't get the right solution. It seems so simple! This makes me wonder if I still miss some basics >-<

0,2g of sodium hydroxide was added to 1,0l of an aqueous solution of the monovalent acid. An addition after, the pH of the solution was 3,00. How many moles of acide were dissolved in the water before addition of sodium hydroxide? The pKa of the acid is 4,00. The addition of sodium does not change colume of the solution (t= 25°C) The correct answer is: 0,066mol

[exchemist]

22 hours ago, Nigthwishxd said:

Hi! I have tried to solve this for few days but can't get the right solution. It seems so simple! This makes me wonder if I still miss some basics >-<

0,2g of sodium hydroxide was added to 1,0l of an aqueous solution of the monovalent acid. An addition after, the pH of the solution was 3,00. How many moles of acide were dissolved in the water before addition of sodium hydroxide? The pKa of the acid is 4,00. The addition of sodium does not change colume of the solution (t= 25°C) The correct answer is: 0,066mol

Where have you got to before you get stuck?

[Nigthwishxd]

17 hours ago, exchemist said:

Where have you got to before you get stuck?

Hey, thanks for looking into this!

n(NaOH)= 0,2g/39,99= 0,005 mol

Ka =10^-4

End [H3O] =10^-3

Equillibrium after adding base, I have no idea how this should be formulated to solve [HA] in the beginning.

OH- + HA --> A- + H2O 

10^-4= (10^-3)x(X+0,005)/(X-0,005)

 

[exchemist]

4 hours ago, Nigthwishxd said:

Hey, thanks for looking into this!

n(NaOH)= 0,2g/39,99= 0,005 mol

Ka =10^-4

End [H3O] =10^-3

Equillibrium after adding base, I have no idea how this should be formulated to solve [HA] in the beginning.

OH- + HA --> A- + H2O 

10^-4= (10^-3)x(X+0,005)/(X-0,005)

 

It's a long time since I did these, but I suspect the key is the equilibrium constant for the dissociation of the acid. NaOH is a strong base so it will mop up essentially all the H+ it can react with, leaving you with a lot more A- than H+. So that will tell you how much extra A- you have, and you know how much H+ there is from the pH at the end.  So by plugging these into the formula for the equilibrium constant, I think you have enough information to construct an equation to solve for the unknown, don't you? But I admit I have not yet got out pen and paper and tried.