Calculate the percent ionization of nitrous acid in a solution that is 0.210 mol L^-1 in nitrous acid and 0.290 mol L^-1 in potassium nitrite ( ). The acid dissociation constant of nitrous acid is A) 58.0 B) 0.154 C) 16.6 D) 2.74 × E) 1.55

Question

Quizplus · Accepted Answer

First, we need to find the concentration of nitrous acid that has dissociated. Let x be the concentration of HNO2 that has dissociated. Then, the concentration of HNO2 that hasn't dissociated is 0.210 - x. The concentration of NO2- (from the dissociation) is also x. Now we can use the acid dissociation constant to write the expression for the equilibrium constant: Ka = [H+][NO2-]/[HNO2] Substituting in the concentrations we just found and simplifying, we get: Ka = x^2 / (0.210-x) The percent ionization is equal to the concentration of dissociated HNO2 divided by the initial concentration of HNO2, multiplied by 100%. In other words: % ionization = (x / 0.210) * 100% We can solve for x by using the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a where a = 1, b = 0, and c = -Ka*0.210. Using the given value of Ka, we find: x = 0.0341 mol L^-1 So, the percent ionization is: % ionization = (0.0341 / 0.210) * 100% = 16.2% This is closest to answer choice B (0.154), which is the best choice.

Calculate the Percent Ionization of Nitrous Acid in a Solution