Volume problem?

Question

What volumes of 0.41 M HNO2 and 0.57 M NaNO2 must be mixed to prepare 1.00 L of a solution buffered at pH = 3.64?

Answer 1

This is a buffer problem, so the Henderson-Hasselbach equation is the tool to use:

pH=pKa +log[conj.base]/[acid] = pKa + log[NO2(-)]/[HNO2]

For HNO2 pKa=3.34
Let's assume we need V1 L of HNO2 and V2 L of NaNO2.
Then V1+V2=1

Using this equation we frequently do the approximation that the equilibrium concentrations are practically the same as the initial ones (if you want higher accuracy you ought to set up an ICE table). We will do this approximation, thus we have to find the initial concentratios just after mixing the two solutions.

[HNO2]= mole/V= M1*V1/Vtotal
[NO2(-)]= [NaNO2]= M2V2/Vtotal
So [NO2(-)]/[HNO2] = M2V2/(M1V1)
substitute in the Hend-Has. Equation

pH=pKa + log(M2V2/M1V1) =>
3.64 = 3.34 + log(0.57V2/0.41V1) =>
log(0.57V2/0.41V1) =0.3 =>
0.57V2/0.41V1 =10^0.3 =>
V2=(0.41/0.57)*(10^0.3) *V1 =1.44 V1

But V1+V2=1 => V1+1.44V1=1 =>
V1 =1/1.44 = 0.694 L= 694 mL HNO2
and V2= 1-V1= 1-0.694 =0.306 L =306 mL NaNO2