Quantum Decibels

A Neutron star is the only thing in the universe not composed of chemicals. It is composed of, as the name implies, a soup of ultra dense neutrons, and would not resemble any recognizable form of matter.

….In honor of my completed chemistry final….

What is not composed of chemicals?

Question (for reference): 

Determine the number of moles of HNO2 and NaNO2 in their buffer solution given the following information:

.44 M HNO2                               

.53 M NaNO2                              

pH=3.59

pKa=3.3

Total volume is 1.00 L

Answer:

 To find the number of moles of each species we must first calculate the volume of each species. Using the Henderson-Hasselbach we can obtain an expression for the ration of equilibrium molarities of the two. ( Substituting A for the base NO2, and HA for the acid: H+)

pH = pKa + log([A] / [HA])

Solving for [A] / [HA] gives us:

[A]/[HA] = 10^(pH-pKa)

this, by plugging in pH and pKa, is equivalent to

[A]/[HA] = 10^(.29)

Now, these two concentrations, [A] and [HA] are not the same as the given original concentrations, so we cannot simply solve. Instead they are the equilibrium concentrations.

This means that this is equal to the moles of A and HA over the total volume:

[A] = mol A / Vt

and

[HA] = mol HA / Vt

where Vt is the total volume, or 1.00L. As it is 1L we can neglect it and simply write:

[A] = mol A

[HA] = mol HA

From this we obtain the mole ratio of the species or:

10^(.29) = mol A / mol HA

However, both of these are unknowns, so we must express one in terms of the other:

mol A = mol HA * 10^(.29)

Now that we have this in terms of one variable, we can use this to solve for the individual volumes of A and HA that will give us the resultant pH using the formula:

Molarity = moles / volume

thus:

.44 = mol A / Va

and

.53 = mol HA / Vha

Where Va and Vha are the volume contributed by A and HA, respectively.

Here however, we have four unknowns. But we can eliminate mol A as we know that:

mol A = mol HA * 10^(.29)

thus:

.44 = (mol HA * 10^(.29)) / Va

We can now solve for Va in terms of HA:

Va = (mol HA * 10^(.29)) / .44

Also solving for Vha in the other equation gives us:

Vha = mol HA / .53

Now, we know that the total volume is 1 L, thus:

Vha + Va = 1

Therefore:

{(mol HA * 10^(.29)) / .44} + {mol HA / .53} = 1

Giving us, when we solve, that:

mol HA = .1583 moles

Thus:

mol A = mol HA * 10^(.29)

and:

mol A = .3086 moles

or:

mol HNO2 = .1583 moles

and:

mol NaNO2 = .3086 moles

Plugging these numbers back into the Henderson-Hasselbach equation we get that:

pH = 3.3 + log(.3086 / .1583)

or that:

pH = 3.59

Thus our answer is indeed correct.

To solve for the initial volume simply use:

V = mol / Molarity

This gives:

Va ~ .4 L

and

Vha ~ .6 L 

What are the number of moles and what are the volumes of the initial solutions of .44 M HNO2 and .53 M NaNO2, if when combined they form a buffer solution of pH 3.59 and volume 1 L?

Info:

.44 M HNO2                               

.53 M NaNO2                              

pH=3.59

pKa=3.3

Total volume is 1.00 L

(Answer will be posted in a few days)

demagogue \DEM-uh-gog, noun:
1. A leader who obtains power by means of impassioned appeals to the emotions and prejudices of the populace.
2. A leader of the common people in ancient times.

“…No more are the days that I will 
Fear, for I have found a strength that 
None can match and I’ll push forward 

Never has the blood in my veins 
Flowed so fiercely as when I feel this around me 
I am whole 

I’m not alone 
With the touch of your hand 
I am whole again…”

(All that Remains)