Question

A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0 mL volumetric flask and diluting to the calibration mark. If 14.47 mL of a 1.23 M solution of potassium hydroxide is added to the buffer, what is the final pH? The Ka for nitrous acid = 4.6 X 10-4.

Answer 1

the buffer:

pH = pKa + log(NaNO2/HNO2)

pKa = -logKa = -log(4.6*10^-4) = 3.33724

then

initially:

NaNO2 = masS/MW = 5.86/68.9953 = 0.0849 mol initially

then

mol of HNO2 = MV = 32.6*4.9 =159.74 mmol = 0.15974 mol of HNO2

after addition o f

KOH = MV = 14.47*1.23 = 17.7981 mmol of base = 17.7981*10^-3 mol of base

then

NO2- forms = 0.0849 + 17.7981*10^-3 = 0.1026981

HNO2 reacts = 0.15974 - 17.7981*10^-3 = 0.1419419

finally

pH = 3.33724+ log(0.1026981/0.1419419)

pH = 3.19669