Question

Consider the titration of 25mL of 0.10 M HNO2 with 0.15 M NaOH. The Ka of nitrous acid is 4.0 X 10^-4.

a) Write the balanced equation for the neutralization reaction.

b) What is the pH of the acid solution before addition of any NaOH?

c) How many milliliters of NaOH are required to reach the equivalence point?

d) What is the pH at the halfway point of the titration?

e) Will the pH at the equivalence be less than 7, equal to 7, or greater than 7?

Answer 1

a) HNO2 + NaOH <------> NaNO2 + H2O

b) HNO2 <-------> H+ + NO2-

0.1___________0_____0

0.1-x__________x______x

Ka= [H+][NO2-]/[HNO2]= x²/0.1-x= 4.0x10^-4

x= 6.13x10^-3M= [H+] -----> pH= -log[H+] ---> pH= 2.21

c) mol of acid= 0.1M x 0.025L= 2.5x10^-3 mol

we need equal moles of NaOH to reach the equivalence point:

Volume NaOH= 2.5x10^-3 mol/0.15M= 0.0167L= 16.7 mL

d) pH at halfway point= pKa

pKa= -log Ka= 3.4= pH

e) it will be bigger than 7 becasu the salt NaNO2 will dissociate to give Na+ and NO2-. The Na+ is the conjugate acid of a strong base, so it is a really weak acid and won´t affect the pH of the final solution. But the NO2- is the conjugate base of a weak acid, so it will be a strong base:

NO2- + H2O <------> HNO2 + OH-

That OH- formed affect the pH, we will have a basic soluion at the end, pH will be >7.