- #1
ksinclair13
- 99
- 0
Wow, I am having a brain fart :(. I can't seem to figure this question out.
Here's some information that I've figured out:
[tex]pH = pKa + log (\frac{[NO_2^-]}{[HNO_2]})[/tex]
[tex]3.00 = 3.40 + log (\frac{[NO_2^-]}{[HNO_2]})[/tex]
[tex]\frac{[NO_2^-]}{[HNO_2]} = .40[/tex]
Also,
[tex]pH = -log[H^+][/tex]
[tex][H^+] = .0010 M[/tex]
I've made an "ICE" chart as well (Initial, Change, Equilibrium) using this equation for the set up:
[tex]HNO_2 \longrightarrow H^+ + NO_2^-[/tex]
I also know that the amount of mL of [itex]HNO_2[/itex] is 1000 mL - mL of [itex]NO_2^-[/itex].
Finally, I know that answer is 715 mL [itex]HNO_2[/itex] and 285 mL [itex]NaNO_2[/itex]. 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 :).
You have solutions of .100 M [itex]HNO_2[/itex] and .100 M [itex]NaNO_2[/itex] ([itex]K_a[/itex] for [itex]HNO_2 = 4.0 x 10^{-4}[/itex]). 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:
[tex]pH = pKa + log (\frac{[NO_2^-]}{[HNO_2]})[/tex]
[tex]3.00 = 3.40 + log (\frac{[NO_2^-]}{[HNO_2]})[/tex]
[tex]\frac{[NO_2^-]}{[HNO_2]} = .40[/tex]
Also,
[tex]pH = -log[H^+][/tex]
[tex][H^+] = .0010 M[/tex]
I've made an "ICE" chart as well (Initial, Change, Equilibrium) using this equation for the set up:
[tex]HNO_2 \longrightarrow H^+ + NO_2^-[/tex]
I also know that the amount of mL of [itex]HNO_2[/itex] is 1000 mL - mL of [itex]NO_2^-[/itex].
Finally, I know that answer is 715 mL [itex]HNO_2[/itex] and 285 mL [itex]NaNO_2[/itex]. 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 :).
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