# Question 15.3: If you mix equal concentrations of reactants and products, w......

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

STRATEGY

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 Stronger acid Acid, HA Base, $A^{-}$ Weaker base $\left.\begin{matrix}HClO_{4}\\HCl\\H_{2}SO_{4}\\HNO_{3}\end{matrix}\right\}$ Strong acids: 100% dissociated in aqueous solution. $\left.\begin{matrix}ClO_{4} ^{-}\\Cl^{-}\\HSO_{4} ^{-}\\NO_{3} ^{-}\end{matrix}\right\}$ 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. $H_{2}O$ $OH^{-}$ Weaker acid $\left.\begin{matrix}NH_{3}\\OH^{-}\\H_{2}\end{matrix}\right\}$ 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. Stronger base
Step-by-Step
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(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.

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

Question: 15.1

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Question: 15.14

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Question: 15.12

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Question: 15.11

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Question: 15.10

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Question: 15.2

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Question: 15.15

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Question: 15.13

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Question: 15.9