Question

What volume of .49 m hno2 and o.49m nano2 must be mixed to prepare 1 l of sodium buffered at ph 2.96?

Answer 1

pKa = - log 4.0 x 10^-4 = 3.40 
pH = pKa + log [NO2-] / [HNO2] 
3.55 = 3.40 + log [NO2-] / [ HNO3] 
0.15 = log [NO2- ] / [HNO2] 
[NO2-] / [HNO2] = 10^0.15 = 1.41 
the ratio between NO2- and HNO2 is 1.41 
This means 1.41 of NO2- / 1.0 HNO2 
1.0 L / 2.41 = 0.415 
Volume of HNO3 = 0.415 L 
Volume of NO2- = 1 - 0.415 = 0.585 L