ICE tables Chemistry

Question

Determine concentration of hydroxide ion, OH ^- , in a solution of HNO2 by constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the concentration of OH ^ - . The Ka for HNO2 is 7.1x10^-4.

A 0.60 M aqueous solution of HNO2 is prepared. Fill in the ICE table with the appropriate values for each involved species to determine concentration of all reactants and products.

HNO2 (aq) + H2O(l) = H3O+ + NO2^-

Ka = ........ = 7.1 x10^-4

[OH-]= M

Answer 1

Starting with the balanced equation:

HNO₂(aq) + H₂O(l) <==> H₃O⁺(aq) + NO₂^-(aq)

0.60....................................0.................0...............Initial

-x......................................+x................+x..............Change

0.60-x................................x...................x..............Equilibrium

Next, write the Ka expression:

Ka = [H₃O⁺][NO₂^-] / [HNO₂]

Now, fill in the values from the ICE table, and solve for x:

7.1x10^-4 = (x)(x) / 0.60 - x (because Ka is small, we can ignore x in the denominator and simplify)

7.1x10^-4 = (x)(x) / 0.60

x² = 4.26x10^-4

x = 0.0206 = [H₃O⁺] = [NO₂^-] (note that 0.0206 is only ~ 3% of 0.60 so above assumption was valid)

The question asks to find the [OH-], so we use the equation for the autoionization of H2O:

Kw = [H₃O⁺][OH^-] = 1x10^-14

1x10^-14 = [0.0206][OH^-]

[OH^-] = 1x10^-14 / 0.0206

[OH^-] = 4.9x10^-13 M