Question

What volume (to the nearest 0.1 mL) of 4.00-M NaOH must be added to 0.650 L of 0.350-M HNO2 to prepare a pH = 3.30 buffer?

Answer 1

Let volume of NaOH be V L

mol of NaOH added = 4*V mol

Before adding NaOH

Before Reaction:

mol of NO2- = 0 mol

mol of HNO2 = 0.35 M *0.65 L

mol of HNO2 = 0.2275 mol

4*V NaOH will react with 4*V of HNO2 to form extra 4*V of base

After adding NaOH

mol of HNO2 = 0.2275-4*V mol

mol of NO2- = 0+4*V mol

Ka = 4.5*10^-4

pKa = - log (Ka)

= - log(4.5*10^-4)

= 3.3468

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

3.3 = 3.3468+log {[NO2-]/[HNO2]}

log {[NO2-]/[HNO2]} = -0.0468

[NO2-]/[HNO2] = 0.8979

So,

(4*V)/(0.2275-4*V) = 0.8979

4*V = 0.2043 - 3.5915*V

(4+3.5915)*V = 0.2043

V = 0.0269 L

V = 26.9 mL

Answer: 26.9 mL