# 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. 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         ...

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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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