Exercise 9 | Buffers and Acid-Base Titrations

General Chemistry 3 - Exercise 9

Calculate the mass of NaNO2 (s) that must be added to 1.00 L of 0.500 M HNO2 (aq) to give a pH of 3.70


Data: Ka (HNO2/NO2-) = 5.62 x 10-4

Henderson-Hasselbalch equation can be used:

pH = pKa + log [base]0acid0

pH = 3.25 + log NO2-0HNO20

NO2-0HNO20 = 100.45

 

NO2-0HNO20= nNO2-V × HNO20 = mNaNO2MNaNO2 × V × HNO20

⇒ mNaNO2 = 100.45 x MNaNO2 V x [HNO2]0

⇒ mNaNO2 = 97.2 g