Respuesta :
The percent ionization in the nitrous acid solution is 1.97% before, and 0.0999% after the addition of sodium nitrate, which lowered the ionization 19.7 times.
To calculate percent ionization, we need to use the expression for the acidity constant of nitrous acid:
[tex]Ka = \frac{[H^{+} ][NO_{2} ^{-} ]}{[HNO_{2} ]}[/tex]
Ka - acidity constant (for nitrous acid, 7.1 * 10⁻⁴)
[H⁺] - concentration of H⁺ ions
[NO₂⁻] - concentration of nitrite ions
[HNO₂] - concentration of unionized nitrous acid
[H⁺] = [NO₂⁻]
[HNO₂] = 1.80 M - X
[tex]7.1 * 10^{-4} = \frac{X^{2} }{1.80 - X}[/tex]
When we transform the equation above, we get a quadratic equation:
X² + 7.1 * 10⁻⁴X - 12.78 * 10⁻⁴ = 0
Once we solve for X, we get:
X = 0.0354 M
Because the concentration of H⁺ ions (X) is equal to the concentration of the ionized nitrous acid, the percent ionization will be:
%ion = 100% * X / 1.80 M
%ion = 100% * 0.0354 M / 1.80 M
%ion = 1.97%
The second calculation is actually easier because we can safely assume that, with added sodium nitrite, all nitrite ions come from the salt, and the amount coming from the acid is negligible. With that in mind, the expression for the acidity constant will look like this:
[tex]7.1 * 10^{-4} = \frac{X * 0.71 }{1.80 M - X}[/tex]
When rearranged, we get this equation:
0.71 * X = 12.78 * 10⁻⁴ M - 7.1 * 10⁻⁴X
0.71071 X = 12.78 * 10⁻⁴ M
X = 12.78 * 10⁻⁴ M / 0.71071
X = 0.001798 M
%ion = 100% * 0.001798 M / 1.80 M
%ion = 0.0999%
So the addition of sodium nitrite resulted in 1.97 / 0.0999 = 19.7 times lower ionization of the nitrous acid.
You can learn more about percent ionization here:
brainly.com/question/13161344
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