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Question

What volumes of 0.58 M HNO2 and 0.59 M NaNO2 must be mixed to prepare 1.00 L of a solution buffered at pH = 3.46?
HNO2

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 concentrations just after mixing the two solutions.

[HNO2]= mole/V= M1*V1/Vtotal
[NO2(-)]= [NaNO2]= M2V2/Vtotal
So [NO2(-)]/[HNO2] = M2V2/(M1V1)

substitute in the Henderson-Hasselbalch equation

pH=pKa + log(M2V2/M1V1) =>
3.46 = 3.34 + log(0.59V2/0.58V1) =>
log(0.59V2/0.58V1) =0.12 =>
0.59V2/0.58V1 =10^0.12 =>
V2=(0.58/0.59)*(10^0.12) *V1 = 1.3*V1

But V1+V2=1 => V1+1.3V1=1 =>
V1 =1/2.3 = 0.435 L= 435 mL HNO2
and V2= 1-V1= 1-0.435=0.565 L =565 mL NaNO2

Answer 2

Most people probably don't help you because they either don't know or think that it's homework and you should do it on your own.

Answer 3

yeah, i would like to help, considering i took chemistry like two years ago, but i really hated it, and once i graduated, all that chem poop went out my soul!!!!!!!!!!!! lol have you heard of the teacher? she might be able to help, or tutoring...

Answer 4

i can yelp?
the volume is max and i mummy tape

p=ache &^%?

h no 2

Answer 5

I'm really sorry but we just don't know.