# Video: Applying Knowledge of the Equilibria, and the Effect of Adding Sodium Hydrogen Carbonate on the Reaction of HCO₃⁻ and H₂O

For statements I and II, state for each if they are true or false. I) Adding some sodium hydrogen carbonate, NaHCO₃, to the equilibrium mixture represented by the equation HCO₃⁻ + H₂O ⇄ CO₃²⁻ + H₃O⁺ will decrease the concentration of CO₃²⁻. II) The equilibrium constant of a reaction changes as the concentration of the products changes. If both the true, state if II is a correct explanation for I.

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### Video Transcript

For statements I and II, state for each if they are true or false. I) Adding some sodium hydrogen carbonate, NaHCO₃, to the equilibrium mixture represented by the equation HCO₃⁻ plus H₂O in a reversible reaction with CO₃²⁻ plus H₃O⁺ will decrease the concentration of CO₃²⁻. II) The equilibrium constant of a reaction changes as the concentration of the products changes. If both the true, state if II is a correct explanation for I.

Let’s start with statement I. Sodium hydrogen carbonate is a highly soluble sodium salt. If we add sodium hydrogen carbonate to pure water, it would dissociate into sodium ions and hydrogen carbonate ions. So adding sodium hydrogen carbonate to our equilibrium mixture will probably raise the concentration of hydrogen carbonate ions. We can use Le Chatelier’s principle to predict what should happen in this case.

Le Chatelier’s principle tells us that for a dynamic equilibrium, when the conditions change, the position of equilibrium will shift to counteract the change. In this case, our change is an increase in the concentration of the hydrogen carbonate anion. In our equilibrium, the hydrogen carbonate anion is reacting with water to form the carbonate anion and a hydronium ion. So the forward reaction reduces the concentration of the hydrogen carbonate anion.

So adding sodium hydrogen carbonate will favor the forward reaction since that’s the reaction that’s going to counter our change. However, the forward reaction increases the concentration of carbonate anions. So instead of decrease in the concentration of the carbonate anion, adding sodium hydrogen carbonate to our mixture will result in an increase in its concentration. This is because the more sodium hydrogen carbonate we have, the more hydrogen carbonate anions we have, the faster the forward reaction will occur, and the system will equilibrate closer to the products than before. So statement I is definitely false.

Statement II says that the equilibrium constant of a reaction changes as the concentration of the products changes. The equilibrium constant for an equilibrium is usually given the symbol Kc. For this equilibrium, the equilibrium constant might have this expression: the concentration of carbonate multiplied by the concentration of hydronium divided by the concentration of the hydrogen carbonate anion. Here, we’re assuming that water in our equilibrium is the solvent. So we ignore it.

You can think of the equilibrium constant like an indicator for the final balance between the products and the reactants once we reach equilibrium. The equilibrium constant is essentially fixed for a given equilibrium at a given temperature. You can think of the equilibrium constant like the destination of a journey. The system always has the same equilibrium constant, but it may take different routes to get there. So even if we had a high concentration of any given product, the system would simply compensate, for instance, by having a higher concentration of reactant.

The equilibrium constant — the clue is in the name — is constant. So statement II is also false. Since both statement A and B are false, we don’t have to address the last part of the question. A false statement cannot explain a false statement.