Chapter 15

Q. 15.3

If you mix equal concentrations of reactants and products, which of the following reactions proceed to the right and which proceed to the left?

(a) H_{2}SO_{4}(aq) + NH_{3}(aq)\xrightleftharpoons{}NH_{4}  ^{+}(aq) + HSO_{4}  ^{-} (aq)

(b) HCO_{3}  ^{-} (aq) + SO_{4}  ^{2-} (aq)\xrightleftharpoons{}HSO_{4}  ^{-} (aq) + CO_{3}  ^{2-} (aq)


To predict the direction of reaction, use the balanced equation to identify the acids and bases, and then use Table 15.1 to identify the stronger acid and the stronger base. When equal concentrations of reactants and products are present, proton transfer always occurs from the stronger acid to the stronger base.

TABLE 15.1 Relative Strengths of Conjugate Acid–Base Pairs
Acid, HA Base, A^{-}  Weaker
\left.\begin{matrix}HClO_{4}\\HCl\\H_{2}SO_{4}\\HNO_{3}\end{matrix}\right\}  Strong acids:
100% dissociated
in aqueous
\left.\begin{matrix}ClO_{4}  ^{-}\\Cl^{-}\\HSO_{4}  ^{-}\\NO_{3}  ^{-}\end{matrix}\right\} Very weak bases:
Negligible tendency
to be protonated in
aqueous solution.
H_{3}O^{+} H_{2}O
\left.\begin{matrix}HSO_{4}  ^{-}\\H_{3}PO_{4}\\HNO_{2}\\HF\\CH_{3}CO_{2}H\\H_{2}CO_{3}\\H_{2}S\\NH_{4}  ^{+}\\HCN\\HCO_{3}  ^{-}\end{matrix}\right\}  Weak acids:
Exist in solution
as a mixture of
HA, A^{-}, and H_{3}O^{+}.
\left.\begin{matrix}SO_{4}  ^{2-}\\H_{2}PO_{4}  ^{-}\\NO_{2}  ^{-}\\F^{-}\\CH_{3}CO_{2}  ^{-}\\HCO_{3}  ^{-}\\HS^{-}\\NH_{3}\\CN^{-}\\CO_{3}  ^{2-}\end{matrix}\right\}  Weak bases:
Moderate tendency
to be protonated in
aqueous solution.
H_{2}O OH^{-}
\left.\begin{matrix}NH_{3}\\OH^{-}\\H_{2}\end{matrix}\right\} Very weak acids:
Negligible tendency
to dissociate.
\left.\begin{matrix}NH_{2}  ^{-}\\O^{2-}\\H^{-}\end{matrix}\right\} Strong bases:
100% protonated in
aqueous solution.


Verified Solution

(a) In this reaction, H_{2}SO_{4} and NH_{4}  ^{+} are the acids, and NH_{3} and HSO_{4}  ^{-} are the bases. According to Table 15.1, H_{2}SO_{4} is a stronger acid than NH_{4}  ^{+} and NH_{3} is a stronger base than HSO_{4}  ^{-}. Therefore, NH_{3} gets the proton and the reaction proceeds from left to right.

\underset{Stronger  acid}{H_{2}SO_{4}(aq)}+\underset{Stronge r base}{NH_{3}(aq)}\longrightarrow \underset{Weaker  acid}{NH_{4}  ^{+}(aq)}+\underset{Weaker  base}{HSO_{4}  ^{-}(aq)}

(b) HCO_{3}  ^{-} and HSO_{4}  ^{-} are the acids, and SO_{4}  ^{2-} and CO_{3}  ^{2-} are the bases. Table 15.1 indicates that HSO_{4}  ^{-} is the stronger acid and CO_{3}  ^{2-} is the stronger base. Therefore, CO_{3}  ^{2-} gets the proton and the reaction proceeds from right to left.

\underset{Weaker  acid}{HCO_{3}  ^{-}(aq)}+\underset{Weaker  base}{SO_{4}  ^{2-}(aq)}\longleftarrow\underset{Stronger  acid}{HSO_{4}  ^{-}(aq)}+\underset{Stronger  base}{CO_{3}  ^{2-}(aq)}