calculate the pH during titration??????

Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid:
a) 0 mL?
b)10.00 mL?
im having problems with the steps so if you could show the steps that would be great!

KOH and HBr are strong a strong base and acid. so i don't have a Ka or Kb

Answer 1

a) KOH is a strong electrolyte and it's assumed to be 100% dissociated.So this means that 0.1000 M KOH gives 0.1000 M K+ and 0.1000 M OH-.
Once we know the concentration of OH- we can calculate H+ from Kw
H+ = 1 10^-14/0.1000= 1 10^-13
pH=13
b) I Have not Ka of HBr. Can you be so kind to tell me and i can answer

Answer 2

You need to figure out the total volume at each stage and determine what the prevalent species is to figure out the pH.

a) When 0 ml have been added, the pH is simply 14-pOH, where pOH is -log [0.1000] or 1, so pH is 13.
b) After 10 ml has been added, the total volume is now 30 ml. You started with 2 mmol of KOH (20 ml * 0.1000 M) and you added 1 mmol of HBr (10 ml * 0.1000 M), so the prevalent species that is left is still OH-, but now only 1 mmol is left, in 30 ml of solution. 1 mmol/30 ml = 0.0333 M OH-. pOH = -log [0.03333] = 1.48. pH = 14 - 1.48 = 12.52

You can do this at any stage in the titration. Until you've added 20 ml of the acid, the pH will be above 7; after that, it will be below 7. For instance, if you try to calculate pH after the addition of 19 ml of acid, your total volume would now be 39 ml. You will have added 1.9 mmol of acid, so only 0.1 mmol of base will be left as the prevalent species. The concentration would be 0.1/39 or 0.00256 M. pOH would be 2.59, so the pH is 11.41.

Answer 3

Start with the equation and the see the moles of teh acid and salt left after the addition of the base: CH3CH2CH2COOH + NaOH -----> CH3CH2CH2COONa + H2O Start 20.0mL x 0.2000M 21.00mL x 0.1000M 0 4.000 mmol 2.100 mmole Equilibrium 1.900 mmol 0 mmol 2.100mmol Ka = [H+]*[Butanoate] / [butanoic acid] (since the volume is the same the mmol ratio can be used) [H+] = Ka x [butanoic acid] /[Butanoate] = 1.54x10^5- x 1.90/2.1 = 1.39x10^5- pH = -log10[1.39x10^5-] = 4.85