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-.

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

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.

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

The pH at half the equivalence point is equal to the pKa of the acid.

In this question, titration curve would graph the pH of acid solution versus the amount of base added. Since the base is strong and the acid is weak, we can conclude that the pH will be slightly greater than 7 at the equivalence point. The equivalence point is found in the steepest region of the curve.

The half-equivalence point is the flattest region of the titration curve and is most resistant to changes in pH. This corresponds to the pKa of the acid. Within this region, adding base (changing the x-value) results in very little deviation in the pH (the y-value). This region is also the buffer region for the given acid.

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

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.

Recall that oxidation always occurs at the anode (in both the electrochemical and galvanic cells). loses two electrons in this case to become . The presence of is hinted by the ionic compound .