Question

What volumes of .5M HNO2 and .5M NaNO2 must be mixed to prepare 1.00L of a solution buffered at pH=3.51? (Ka for HNO2 = 4X10^-4)

Answer 1

pH= PKa+ log {[A-]/[HA]}

[A-] Conjugate base and HA= HNO2

Ka for HNO2= 4*10^-4

pKa= -log (4*10^-4) =3.39

3.51= 3.39 + log {[A-]/[HA]}

{[A-]/ [HA]} =10^0.12 = 1.32

moles of base/ moles of acid =1.32

Volume of acid + base = 1L since both

Va+ Vb= 1 ( Va and Vb are volume of acid and base)

Volume of base* molarity of base/ Volume of acid* Molarity= 1.32

Volume of base/ Volume of acid = 1.32

Vb/(1-Vb)= 1.32

Vb= 1.32- 1.32Vb

2.32Vb= 1.32

Vb= 1.32/2.32 =0.57 L

Va =1-0.57= 0.43L