Help! Solving for Volumes of HNO_2 and NaNO_2

[ksinclair13]

Wow, I am having a brain fart :(. I can't seem to figure this question out.

You have solutions of .100 M HNO_2 and .100 M NaNO_2 (K_a for HNO_2 = 4.0 x 10^{-4}). What volumes of each are needed to be combined to make 1.00 L of a solution with a pH of 3.00?

Here's some information that I've figured out:

pH = pKa + log (\frac{[NO_2^-]}{[HNO_2]})
3.00 = 3.40 + log (\frac{[NO_2^-]}{[HNO_2]})
\frac{[NO_2^-]}{[HNO_2]} = .40

Also,
pH = -log[H^+]
[H^+] = .0010 M

I've made an "ICE" chart as well (Initial, Change, Equilibrium) using this equation for the set up:

HNO_2 \longrightarrow H^+ + NO_2^-

I also know that the amount of mL of HNO_2 is 1000 mL - mL of NO_2^-.

Finally, I know that answer is 715 mL HNO_2 and 285 mL NaNO_2. I'm sure I could come up with the answer if I spent a while at it - I know I'm close. But it would be great if someone helped me out a little bit :).

 

[Borek]

No need for ICE, assume there is no reaction taking place - just the ratio given by H-H equation must be maintained. So, x+y=1L, x/y=0.4, that's all.

 

[GCT]

adding to what borek said

x=moles of nitrous acid, y=moles of conjugate base

(x/y)=.4, y=x/.4

x(1 L /.100 moles HNO2) + (x/.4)(1 L/ .100 moles NO2)=1 L

now find the x and y values and you should be able to go from there

 

[ksinclair13]

Wow, that was much easier than I thought. Thank you very much for the help :). It would've taken me a long time to figure that out because I assumed you had to make an "ICE" chart.

 

[GCT]

You should read up on the henderson-hasselbach equation, you can use it in the buffer regions usually in the 10th molarity ratios with respect to both substances (that is one substance is 10 times higher in concentration then the other), at these regions, it's usually safe to say that neither of the compounds go on to react with water, but each of the components, somewhat suppresses such reactions, since they are mutual products (Think La Chatelier's Principle)