Question

In a titration of 100 mL 0.10 M Nitrous acid HNO₂ (Ka= 4.5*10^-4) against a 0.10 M Sodium hydroxide NaOH, calculate the pH at the equivalence point?

Answer 1

To solve the problem we need to know the number of moles of acid:

Moles = M x L = (0,10M)(0,100 L) = 0,0100 mol

From VaMa=VbMb, we know that will take 100 ml of NaOH to reach the equivalence point. So we need to add the total volume (acid (100 ml) and base (100 ml), total 200 ml) to know the NO₂^- concentration:

[NO₂^-] = (0,0100 mol)/(0,200 L) = 0,050 M

The equilibrium reaction of the process will be:

NO₂^- + H₂O <=> HNO₂ + OH^-

The Kb for the NO₂^- can be calculate from the Ka value of its conjugate acid, Kb = Kw/Ka = 1,0x10^-14/4,5x10^-4 = 2,22x10^-11. Using the Kb expression, we have:

Making the approximation that 0,050 - x is almost 0,050, and solving x, we have that x = [OH-] = 1,04x10^-4 , pOH = 5,98 and pH = 8,02