Calculate the percent ionization of nitrous acid in a solution that is 0.230 m in nitrous acid. the acid dissociation constant of nitrous acid is 4.50 ⋅ 10-4.

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        0.230M          0M       0M
Δ[]      -x                   +x         +x
[]f        0.230-x          x           x

Then, using the concentration equation, you get

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

4.5x10^-4 = x*x / .230 - x

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

assume 0.230-x ≈ 0.230

4.5x10^-4 = x^2/0.230

Then, we solve for x by first multiplying both sides by 0.230 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.01M.

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

0.01M/0.230M = .0434 or 

≈4.34% dissociation.

Sugar and water are chemically combined.(a solution is chemically combined while a mixture is mixed together without bonding the molecules)