[flagpolesitta43]

My first exam for Chem 2 is coming up this week and I'm not sure if I'm solving some of these buffer solution problems correctly. So if someone could let me know if I'm doing it correctly or not it would be a really big help. Thanks.

You want to make a buffer solution of pH=3.5. How many grams of sodium nitrite should be added to a .500L sample of .10M nitrous acid? You may neglect the volume of solid acid added.

First I wrote out my Ka expression: Ka=[NO2-][H30+]/[HNO2]

I can get the concentration of [H30+] from the given pH of 3.5, which will equal 3.2x10^-4M

Then I just plugged in my numbers: 4.5x10^-4=

• [3.2x10-4M]/[.10M]

Which will give you [.14M NO2-].

So to get to grams I: (.14M)(.500L)(69g/mol)=4.83g of NaNO2 needed to make the buffer.

Is any of this right?

[Albert]

It seems you solved it in the right way. However, you had 4.5x10^-4 as Ka, while I've always used 7.1x10^-4 for nitrous acid.

[flagpolesitta43]

Thank-you for replying, maybe you can help me with this problem as well.

This is part B of the question and I'm really not sure what to do:

The pH of a sample of hydrochloric acid is 0.0. What is the pH of the system if 5.00 mL of this acid is added to the .500L buffer you prepared in Part A of this problem? Account for the volume of acid in this problem.

Any help that could be offered would be really appreciated...thanks again.

[Albert]

Hint:

[NaNO2] = [NaNO2] - [H+]

[HNO2] = [HNO2] + [H+]

H+ are those introduced from the 1M solution of HCl.

Moreover: 'forget' about the volumes and put in the Henderson-Hasselbach equation only the moles.

[flagpolesitta43]

My professor does not allow us to use the Henderson-Hasselbach equation, is there another way to solve it?

[Borek]

[NaNO2] = [NaNO2] - [H+]

[HNO2] = [HNO2] + [H+]

H+ are those introduced from the 1M solution of HCl.

You better edit these "equations" before I will take your scooby snack

You use same symbol for analytical concentration and for equilibrium concentration (or rather "quasi-equilibrium" concentration).

flagpolesitta43 - look here: http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-buffer-q2. WHile it is not exactly the same case, it is related - note how amounts of acid and conjugated base are calculated from stoichiometry.

[flagpolesitta43]

Thank you both very much, but I cannot use the Hasselbalch equation, my professor wants it in terms of equlibrium constants. So if you could show me how to solve the problem that way I would really appreicate it.

[Borek]

Start with equilibrium constant and rearrange it to the form of H-H - it takes two steps only.

You may also omit the second step of rearranging (or derivation) and don't use logarithms, thus you will not use H-H equation, ate least from formal point of view:

http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch#eq15.1

[Albert]

Generally speaking

Ka = [H+] [Base-]/[Acid]

Hence: [H+] = Ka [acid]/[base-] = Ka (moles of acid/moles of base)

This is 'derivative' of the HH equation.

Mind you: are you sure about the Ka you've used so far?

[Cation]

Right, knowing the equilibrium constant expression is essential to solving these problems.
My professor routinely makes fun of Henderson and Hasselbach because they got their name
on the freaking formula by taking a base ten log of both sides and calling it theirs.

to answer the question:

1 molar HCl * 5 mL -> 0.005 mol HCl
.14 molar NO3- * 500 mL -> 0.070 mol NO2-
0.1 molar HNO2 * 500 mL -> 0.05 mol HNO2

The equation is:

H3O+ + NO2- -> HNO2 + H2O - so you'll be subtracting moles hcl ( assume complete
dissociation for strong acids ) from no2- and adding the same amount to conjugate
acid HNO2. since everything is in the same volume of solution now you don't even need to
switch back to molarities to solve for h3o+.

[H3O+] = 4.5x10^-4 ( 0.55 mol HNO2 / 0.65 mol NO2- ) = 3.81 x 10^-4

Negative logarithm of that is something like 3.42 = pH. Which is just a little bit below the
pH of the buffer from part a, which makes sense logically. Let me know if that helped / I
missed something!