Nigthwishxd

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)

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