We have to calculate the pH value of a given molarity of a given product.

0.5M solution if NaNO2 is prepared. Calculate the pH of this solution. 100mL of a 2M HCl solution is titrated with a 2M NaOH solution. a)What is the pH of the HCl solution before any of the NaOH solution is added? b) What volume of the NaOH solution is required to reach the equivalence point of the titration? c) What is the pH at the equivalence point of the titration? d) At the end of the titration, the total colume of the NaOH solution added is 200mL, what is the final pH? 100mL of a 2M acetic acid solution is titrated with a 2M NaOH solution. a)What is the pH of the acetic acid solution before any of the NaOH solution is added? b) What volume of the NaOH solution is required to reach the equivalence point of the titration? c) What is the pH at the equivalence point of the titration? d) At the end of the titration, the total colume of the NaOH solution added is 200mL, what is the final pH?

Skill: 

NaNO2 will dissociate into Na+ and NO2-. Sodium ions will not change the pH of water, but nitrite ions are the (weak) conjugate base of weak acid nitrous acid. It will cause the solution to be slightly basic.

H2O + NO2-  ↔  HNO2 + OH-

Kb = [HNO2][OH-]/[NO2-] = Kw/Ka = 1·10-14/7.1·10-4 = 1.41·10-11

NO2- HNO2 OH-

Initial conc. (M) 0.5 0 0

Conc. change (M) -x x x

Final conc. (M) 0.5-x x x

If 0.5>>x, we can approximate 0.5 ≈ 0.5-x

So Kb = 1.41·10-11 = x2/0.5   →  x = [OH-] = 2.65·10-6 

pOH = -log [OH-] = 5.58, so pH = 14 - pOH = 8.42