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Chapter 10: Problem 14

Identify the Br酶nsted-Lowry acid-base pairs in each of the following equations: a. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) b. \(\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NH}_{3}(a q) \rightleftarrows \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q)\)

Short Answer

Expert verified
a: \(\mathrm{H}_{3} \mathrm{PO}_{4} / \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\), \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}_{3} \mathrm{O}^{+}\); b: \(\mathrm{H}_{2} \mathrm{O} / \mathrm{OH}^{-}\), \(\mathrm{CO}_{3}^{2-} / \mathrm{HCO}_{3}^{-}\); c: \(\mathrm{H}_{3} \mathrm{PO}_{4} / \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\), \(\mathrm{NH}_{3} / \mathrm{NH}_{4}^{+}\)

Step by step solution

01

Identifying species in the first equation

Consider the equation \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\). Identify which species donate protons and which species accept protons. Here, \(\mathrm{H}_{3} \mathrm{PO}_{4}\) donates a proton (becomes \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)), and \(\mathrm{H}_{2} \mathrm{O}\) accepts that proton (becomes \(\mathrm{H}_{3} \mathrm{O}^{+}\)).
02

Identify Br酶nsted-Lowry pairs in the first equation

In the equation \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\), the Br酶nsted-Lowry acid-base pairs are: \(\mathrm{H}_{3} \mathrm{PO}_{4} / \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}_{3} \mathrm{O}^{+}\).
03

Identifying species in the second equation

Consider the equation \(\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)\). Identify which species donate protons and which species accept protons. Here, \(\mathrm{H}_{2} \mathrm{O}\) donates a proton (becomes \(\mathrm{OH}^{-}\)), and \(\mathrm{CO}_{3}^{2-}\) accepts that proton (becomes \(\mathrm{HCO}_{3}^{-}\)).
04

Identify Br酶nsted-Lowry pairs in the second equation

In the equation \(\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)\), the Br酶nsted-Lowry acid-base pairs are: \(\mathrm{H}_{2} \mathrm{O} / \mathrm{OH}^{-}\) and \(\mathrm{CO}_{3}^{2-} / \mathrm{HCO}_{3}^{-}\).
05

Identifying species in the third equation

Consider the equation \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NH}_{3}(a q) \rightleftarrows \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q)\). Identify which species donate protons and which species accept protons. Here, \(\mathrm{H}_{3} \mathrm{PO}_{4}\) donates a proton (becomes \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)), and \(\mathrm{NH}_{3}\) accepts that proton (becomes \(\mathrm{NH}_{4}^{+}\)).
06

Identify Br酶nsted-Lowry pairs in the third equation

In the equation \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NH}_{3}(a q) \rightleftarrows \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q)\), the Br酶nsted-Lowry acid-base pairs are: \(\mathrm{H}_{3} \mathrm{PO}_{4} / \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{NH}_{3} / \mathrm{NH}_{4}^{+}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proton Donor
In Br酶nsted-Lowry theory, an acid is defined as a **proton donor**. This means it gives up a proton (hydrogen ion, H鈦�) to another substance during a chemical reaction. It's crucial to identify the proton donor in any acid-base reaction to understand how the process works.
For example, in the equation \(\text{H}_{3}\text{PO}_{4}(aq)+\text{H}_{2}\text{O}(l) \rightleftharpoons \text{H}_{2}\text{PO}_{4}^{-}(aq)+\text{H}_{3}\text{O}^{+}(aq)\), phosphoric acid (\(\text{H}_{3}\text{PO}_{4}\)) donates a proton to water (\(\text{H}_{2}\text{O}\)), becoming dihydrogen phosphate (\(\text{H}_{2}\text{PO}_{4}^{-}\)).
Another example is the equation \(\text{H}_{3}\text{PO}_{4}(aq)+\text{NH}_{3}(aq) \rightleftharpoons \text{H}_{2}\text{PO}_{4}^{-}(aq)+\text{NH}_{4}^{+}(aq)\). Here, phosphoric acid (\(\text{H}_{3}\text{PO}_{4}\)) again donates a proton, this time to ammonia (\(\text{NH}_{3}\)), forming dihydrogen phosphate (\(\text{H}_{2}\text{PO}_{4}^{-}\)) and ammonium ion (\(\text{NH}_{4}^{+}\)). Identifying the proton donor is the first step to recognize the acid in the reaction.
Proton Acceptor
A **proton acceptor** is a substance that accepts a proton during a chemical reaction. According to Br酶nsted-Lowry theory, a base is defined as a proton acceptor. Knowing which species accepts the proton helps identify the base in the reaction.
For instance, consider the equation \(\text{CO}_{3}^{2-}(aq)+\text{H}_{2}\text{O}(l) \rightleftharpoons \text{HCO}_{3}^{-}(aq)+\text{OH}^{-}(aq)\), carbonate (\(\text{CO}_{3}^{2-}\)) accepts a proton from water (\(\text{H}_{2}\text{O}\)) to form bicarbonate (\(\text{HCO}_{3}^{-}\)). This makes carbonate the proton acceptor, or the base, in the reaction.
Another example is the equation \(\text{H}_{3}\text{PO}_{4}(aq)+\text{NH}_{3}(aq) \rightleftharpoons \text{H}_{2}\text{PO}_{4}^{-}(aq)+\text{NH}_{4}^{+}(aq)\). In this case, ammonia (\(\text{NH}_{3}\)) accepts a proton from phosphoric acid (\(\text{H}_{3}\text{PO}_{4}\)), forming ammonium ion (\(\text{NH}_{4}^{+}\)). Recognizing the proton acceptor helps us identify the base involved in the reaction.
Acid-Base Reactions
An acid-base reaction involves the transfer of a proton from the acid (proton donor) to the base (proton acceptor). Understanding how these reactions work is important for grasping many concepts in chemistry.
Let's look at the reactions provided in the problem:
- In the reaction \(\text{H}_{3}\text{PO}_{4}(aq)+\text{H}_{2}\text{O}(l) \rightleftharpoons \text{H}_{2}\text{PO}_{4}^{-}(aq)+\text{H}_{3}\text{O}^{+}(aq)\), phosphoric acid (\(\text{H}_{3}\text{PO}_{4}\)) donates a proton to water (\(\text{H}_{2}\text{O}\)). In turn, water accepts the proton, becoming hydronium (\(\text{H}_{3}\text{O}^{+}\)). This tells us that phosphoric acid and water are involved in a classic acid-base reaction.
Another example is \(\text{CO}_{3}^{2-}(aq)+\text{H}_{2}\text{O}(l) \rightleftharpoons \text{HCO}_{3}^{-}(aq)+\text{OH}^{-}(aq)\). Here, water (\(\text{H}_{2}\text{O}\)) donates a proton to carbonate (\(\text{CO}_{3}^{2-}\)), forming bicarbonate (\(\text{HCO}_{3}^{-}\)) and hydroxide (\(\text{OH}^{-}\)). These species are also engaged in an acid-base reaction. By recognizing these fundamental processes, you can better understand the dynamics and outcomes of various chemical reactions.

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Most popular questions from this chapter

Which of the following are not at equilibrium? a. The rates of the forward and reverse reactions are equal. b. The rate of the forward reaction does not change. c. The concentrations of reactants and the products are not constant.

Which of the following are at equilibrium? a. The rate of the forward reaction is twice as fast as the rate of the reverse reaction. b. The concentrations of the reactants and the products do not change. c. The rate of the reverse reaction does not change.

Consider the buffer system of hydrofluoric acid, \(\mathrm{HF}\), and its salt, NaF. $$\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q)$$ a. The purpose of this buffer system is to: 1\. maintain [HF] 2\. maintain \(\left[\mathrm{F}^{-}\right]\) 3\. maintain \(\mathrm{pH}\) b. The salt of the weak acid is needed to: 1\. provide the conjugate base 2\. neutralize added \(\mathrm{H}_{3} \mathrm{O}^{+}\) 3\. provide the conjugate acid c. If \(\mathrm{OH}^{-}\) is added, it is neutralized by: 1\. the salt 2\. \(\mathrm{H}_{2} \mathrm{O}\) 3\. \(\mathrm{H}_{3} \mathrm{O}^{+}\) d. When \(\mathrm{H}_{3} \mathrm{O}^{+}\) is added, the equilibrium shifts in the direction of the: 1\. reactants 2\. products 3\. does not change

Why would the pH of your blood plasma decrease if you hold your breath?

State whether each of the following solutions is acidic, basic, or neutral: (10.6) a. saliva, \(\mathrm{pH} 6.8\) b. urine, pH 5.9 c. pancreatic juice, pH 8.0 d. bile, pH 8.4 e. blood, pH 7.45
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