Question

What volume (to the nearest 0.1 mL) of 7.40-M NaOH must be added to 0.400 L of 0.350-M HNO2 to prepare a pH = 3.80 buffer?

Answer 1

For this we have the equation

OH^- + HNO₂ ---> NO₂^- + H₂O

pKa of HNO₂= 3.39 (ref table http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf)

Henderson–Hasselbalch equation:

pH = pKa + log([NO₂^-]/[HNO₂]) concentration of NO₂^- is substituted for concentration of NaOH as amount of NO₂^- is dependent on the amount of NaOH.
3.80 = 3.39 + log(7.40x/ (0.400 L x 0.350 M - 7.40x)) Concentration of HNO₂ is the amount of HNO₂ added subtrated by the amount reacted with NaOH

where x is volume in Liters

3.8 - 3.39 = log(7.40x/ (0.400 L x 0.350 M - 7.40x))

10^0.41 = 7.40x/ 0.14 - 7.40x
(2.57)(0.14 - 7.40x) = 7.40x
0.359 - 19.02x = 7.40x
0.359 = 7.40x + 19.02x

x = 0.359/26.42 = 0.01358 Litres = 13.58 mL