The Ka is the acid dissociation constant, and thus it is what determines how strong the acid is. Stronger acids dissociate to a greater extent and produce lower pH values.

Let's break down the reaction into two separate reactions:

and

We can see that copper loses electrons, while silver gains electrons. Recall that oxidation is loss and reduction is gain, with regard to electrons. Copper is oxidized and silver is reduced.

However, this question asks for the oxidizing agent and reducing agent. Recall that the oxidizing agent is reduced, while the reducing agent is oxidized. Since copper is oxidized, it is the reducing agent. Similarly, since silver is reduced, it is the oxidizing agent.

Hydrogen doesn't change. Cu2+ doesn't change (partnered with S2- then with SO42-). Sulfur goes from S2- and S6+(paired with 6 O2- with a 2– charge), showing an oxidation. Nitrogen goes from N5+ to N2+ meaning it was reduced.

Hydrogen doesn't change. Cu2+ doesn't change (partnered with S2- then with SO42-). Sulfur goes from S2- and S6+(paired with 6 O2- with a 2– charge), showing an oxidation. Nitrogen goes from N5+ to N2+ meaning it was reduced.

The common factor between 2 e- and 5 e- is 10. Therefore the number of electrons involved is 10 e-.

When you balance the redox reaction in acidic conditons, there are 6Fe2+, 1 Cr2O72–, 14 H+, 6 Fe3+, 2 Cr3+, and 7 H2O. Don't forget to add the 1 in front of the Cr2O72–

The Ka is the acid dissociation constant, and thus it is what determines how strong the acid is. Stronger acids dissociate to a greater extent and produce lower pH values.

Since pH > pKA, the deprotonated form of the acid is predominant.

Since is a strong acid, the concentration of is equal to the concentration of the acid itself.

Thus, .

Recall how to find the pH of a solution:

Plug in the given hydronium ion concentration to find the pH of the given solution.

Remember to maintain the correct number of significant figures.

To answer this question we need to look at the reaction below:

An increase in the pH will result in a decrease in the concentration of hydrogen ions (). Using Le Chatelier’s principle we can find out which answer choices will decrease .

Removing carbonic acid will decrease the concentration of . To maintain equilibrium, the reaction will shift to the left and make more reactants from products; therefore, there will be a decrease in the and an increase in pH.

Recall that salts like sodium bicarbonate, or , will dissociate in water and form ions. Sodium bicarbonate will form sodium () and bicarbonate () ions. This side reaction will result in an increase in the bicarbonate ion concentration. Le Chatelier’s principle will shift the equilibrium of the given reaction to the left and, therefore, decrease the . Adding sodium bicarbonate will increase the pH.