Chemistry- Acid and Base question
A 0.150 M solution of nitrous acid (HNO2) is made. Ka = 4.5 x 10-4
1. Show the equilibrium which occurs when this acid is dissolved in water.
2. What is the pH of the solution? Show all work clearly.
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The dissociation is
HNO2 <--> H+ + NO2- (please note that <--> represents the bi-directional equilbrium arrow)
To find the pH of this solution, we need to set up an ICE table as this is a solution at equilibrium.
HNO2 H+ NO2-
I 0.150 M 0 0
C -x +x +x
E 0.150 - x x x
Since Ka is so small, we will make the assumption that x <<0.150 M which means we can ignore that in the mass action expression making the calculations much easier (i.e. no quadratic equation).
We can set up our mass action expression
Ka = [H+][NO2-] / [HNO2]
4.5 x 10^-4 = x^2 / 0.150
We can solve for x to find the [H+] which is 8.2 x 10^-3
Since this is less than 5% of 0.150 M, our assumption was valid.
Now, to find the pH, we need to take the - log[8.2 x 10^-3] to get 2.09.
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