Question

For the following equation: The Ka value of HNO2 is 4.0 x 10-4. If the equilibrium concentration of HNO2 is 0.54 M. what is the equilibrium concentration of NO2/H30? 2 points HNO, + H, O NO + H20+ Your answer

Answer 1

In this question, we are dealing with the dissociation of HNO2, So, let's start by writing the reaction equation with only the relevant species involved in the reaction because water is used in excess and then moving on to the ICE table:

$\boldsymbol{HNO_2 \rightleftharpoons NO_{2}^{-} + H_3O^+}$

Let's draw an ICE table for this reaction:

	HNO2	NO2-	H3O+
Initial Concentration	0.54 M	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.54 - x	x	x

From the given reaction, the equilibrium constant would be:

$\boldsymbol{K_a =\frac{ [NO_{2}^{-}] [H_3O^+]}{[HNO_2]}}$

Let's plug in the values in the expression for the equilibrium constant:

$\boldsymbol{4 \times 10^{-4} =\frac{ x \times x}{(0.54 - x)}}$

$\boldsymbol{4 \times 10^{-4} =\frac{ x^2 }{(0.54 - x)}}$

Solving this quadratic expression would give us the value of x:

$\boldsymbol{x = 0.0145}$

So, the concentration of:

$\boldsymbol{NO_{2}^{-} = 0.0145\: M}$

$\boldsymbol{H_3O^+ = 0.0145\: M}$