Question

Calculate the pH after 0.15 mole of NaOH is added to 1.05 L of a solution that is 0.48 M HNO2 and 1.06 M LiNO2, and calculate the pH after 0.30 mole of HCl is added to 1.05 L of the same solution of HNO2 and LiNO2. ka of hno2 is 4.0 x 10^-4

Answer 1

1)

mol of NaOH added = 0.15 mol

HNO2 will react with OH- to form NO2-

Before Reaction:
mol of NO2- = 1.06 M *1.05 L
mol of NO2- = 1.113 mol

mol of HNO2 = 0.48 M *1.05 L
mol of HNO2 = 0.504 mol

after reaction,
mol of NO2- = mol present initially + mol added
mol of NO2- = (1.113 + 0.15) mol
mol of NO2- = 1.263 mol

mol of HNO2 = mol present initially - mol added
mol of HNO2 = (0.504 - 0.15) mol
mol of HNO2 = 0.354 mol


Ka = 4*10^-4

pKa = - log (Ka)
= - log(4*10^-4)
= 3.398

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.398+ log {1.263/0.354}
= 3.95

Answer: 3.95

2)

mol of HCl added = 0.3 mol

NO2- will react with H+ to form HNO2

Before Reaction:
mol of NO2- = 1.06 M *1.05 L
mol of NO2- = 1.113 mol

mol of HNO2 = 0.48 M *1.05 L
mol of HNO2 = 0.504 mol

after reaction,
mol of NO2- = mol present initially - mol added
mol of NO2- = (1.113 - 0.3) mol
mol of NO2- = 0.813 mol

mol of HNO2 = mol present initially + mol added
mol of HNO2 = (0.504 + 0.3) mol
mol of HNO2 = 0.804 mol


Ka = 4*10^-4

pKa = - log (Ka)
= - log(4*10^-4)
= 3.398

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.398+ log {0.813/0.804}
= 3.403

Answer: 3.40