Question

To a 1.00 L buffer solution made of 1.65 M nitrous acid (HNO2) and 1.01M potassium nitrite (KNO2) was added 0.625 moles of NaOH (assume no volume change to the solution). What is the final pH of this buffer assuming Ka nitrous acid = 4.00e-4

Answer 1

mol of NaOH added = 0.625 mol

HNO2 will react with OH- to form NO2-

Before Reaction:

mol of NO2- = 1.01 M *1.0 L

mol of NO2- = 1.01 mol

mol of HNO2 = 1.65 M *1.0 L

mol of HNO2 = 1.65 mol

after reaction,

mol of NO2- = mol present initially + mol added

mol of NO2- = (1.01 + 0.625) mol

mol of NO2- = 1.635 mol

mol of HNO2 = mol present initially - mol added

mol of HNO2 = (1.65 - 0.625) mol

mol of HNO2 = 1.025 mol

Ka = 4*10^-4

pKa = - log (Ka)

= - log(4*10^-4)

= 3.3979

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

use:

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

= 3.3979+ log {1.635/1.025}

= 3.60

pH is 3.60