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IgorLugansk [536]
3 years ago
15

_and_Are frequently dumped into marine habitats.

Chemistry
1 answer:
Zarrin [17]3 years ago
4 0

Explanation:

if you post the whole question I can help you

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3 years ago
Calculate the percent ionization of nitrous acid in a solution that is 0.139 M in nitrous acid. The acid dissociation constant o
anzhelika [568]
Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

HNO2 <-> H(+) + NO2(-)

Next, create an ICE table

           HNO2   <-->  H+        NO2-
[]i        0.139M          0M       0M
Δ[]      -x                   +x         +x
[]f        0.139-x          x           x

Then, using the concentration equation, you get

4.5x10^-4 = [H+][NO2-]/[HNO2]

4.5x10^-4 = x*x / .139 - x

However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable, 

assume 0.139-x ≈ 0.139

4.5x10^-4 = x^2/0.139

Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.

Then to find percent dissociation, you do final concentration/initial concentration.

0.007M/0.139M = .0503 or 

≈5.03% dissociation.
4 0
3 years ago
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