Calculate the percent ionization of nitrous acid in a solution that is 0.205 M in nitrous acid and 0.295 M in potassium nitrite ( ).The acid dissociation constant of nitrous acid is
A) 59.0
B) 0.151
C) 17.1
D) 2.72 ×
E) 3.508

Question

Calculate the percent ionization of nitrous acid in a solution that is 0.205 M in nitrous acid  and 0.295 M in potassium nitrite (  ).The acid dissociation constant of nitrous acid is 
A) 59.0 
B) 0.151 
C) 17.1 
D) 2.72 × 
E) 3.508

Quizplus · Accepted Answer

Using the given information, we can set up the expression for the acid dissociation constant:
$$K_a=\frac{[	ext{HNO}_2][	ext{NO}_2^-]}{[	ext{H}^+]}$$
We can assume that the concentration of $[	ext{H}^+]$ is equal to the concentration of $[	ext{NO}_2^-]$ because the dissociation of nitrous acid is weak, and the $[	ext{H}^+]$ produced is negligible compared to the initial concentrations of nitrous acid and potassium nitrite. Substituting the given values,
$$1.7	imes10^{-4}=\frac{x(0.295)}{x}$$
where $x$ is the concentration of $[	ext{HNO}_2]$ that dissociates. Solving for $x$ gives $x=4.38	imes10^{-5}$. The percent ionization can be calculated as follows:
$$	ext{percent ionization}=\frac{x}{[	ext{HNO}_2]_{	ext{initial}}}	imes100\%=\frac{4.38	imes10^{-5}}{0.205}	imes100\%\approx0.151\%$$
Therefore, the best choice is B.

Calculate the Percent Ionization of Nitrous Acid in a Solution