This question is incomplete, the complete question is;
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 10.97 mL of a 1.63 M solution of potassium hydroxide is added to the buffer, what is the final pH? The Ka for nitrous acid = 4.6 × 10⁻⁴.
Answer:
the final pH is 3.187
Step-by-step explanation:
Given the data in the question;
Initial moles of HNO2 = 32.6/1000 × 4.90 = 0.15974 mol
Initial moles of NO2- = mass/molar mass = 5.86/68.995 = 0.0849336 mol
Moles of KOH added = 10.97/1000 × 1.63 = 0.0178811 mol
so
HN02 + KOH → NO2- + H2O
moles of HNO2 = 0.15974 - 0.0178811 = 0.1418589 mol
Moles of NO2- = 0.0849336 + 0.0178811 = 0.1028147 mol
Now,
pH = pka + log( [NO2-]/[HNO2])
pH = -log ka + log( moles of NO2- / moles of HNO2 )
we substitute
pH = -log( 4.6 × 10⁻⁴ ) + log( 0.1028147 / 0.1418589 )
pH = -log( 4.6 × 10⁻⁴ ) + log( 0.724767 )
pH = 3.337242 + (-0.1398 )
pH = 3.187
Therefore, the final pH is 3.187