Question

A buffer solution contains 0.341 M nitrous acid and 0.303 M sodium nitrite. If 0.0573 moles of nitric acid are added to 250 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding nitric acid) pH =

Answer 1

mol of HNO3 added = 0.0573 mol

NO2- will react with H+ to form HNO2

Before Reaction:

mol of NO2- = 0.303 M *0.25 L

mol of NO2- = 0.0757 mol

mol of HNO2 = 0.341 M *0.25 L

mol of HNO2 = 0.0853 mol

after reaction,

mol of NO2- = mol present initially - mol added

mol of NO2- = (0.0757 - 0.0573) mol

mol of NO2- = 0.0185 mol

mol of HNO2 = mol present initially + mol added

mol of HNO2 = (0.0853 + 0.0573) mol

mol of HNO2 = 0.1426 mol

Ka = 4.5*10^-4

pKa = - log (Ka)

= - log(4.5*10^-4)

= 3.347

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.347+ log {1.845*10^-2/0.1426}

= 2.459

Answer: 2.46