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Chapter 16: Problem 17

(a) Give the conjugate base of the following BronstedLowry acids: (i) \(\mathrm{HIO}_{3}\), (ii) \(\mathrm{NH}_{4}{ }^{+}\) (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{O}^{2-}\), (ii) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\).

Short Answer

Expert verified
(a) (i) Conjugate base of \(\mathrm{HIO}_{3}\) is \(\mathrm{IO}_{3}^-\). (ii) Conjugate base of \(\mathrm{NH}_{4}^{+}\) is \(\mathrm{NH}_{3}\). (b) (i) Conjugate acid of \(\mathrm{O}^{2-}\) is \(\mathrm{OH}^{-}\). (ii) Conjugate acid of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is \(\mathrm{H}_{3} \mathrm{PO}_{4}\).

Step by step solution

01

(i) Conjugate base of \(\mathrm{HIO}_{3}\)

To find the conjugate base of \(\mathrm{HIO}_{3}\), we need to remove a proton (H鈦�) from its structure to form the conjugate base: \[ \mathrm{HIO}_{3} \rightarrow \mathrm{H}^+ + \mathrm{IO}_{3}^-\] So the conjugate base of \(\mathrm{HIO}_{3}\) is \(\mathrm{IO}_{3}^-\).
02

(ii) Conjugate base of \(\mathrm{NH}_{4}^{+}\)

To find the conjugate base of \(\mathrm{NH}_{4}^{+}\), we need to remove a proton (H鈦�) from its structure to form the conjugate base: \[ \mathrm{NH}_{4}^{+} \rightarrow \mathrm{H}^+ + \mathrm{NH}_{3}\] So the conjugate base of \(\mathrm{NH}_{4}^{+}\) is \(\mathrm{NH}_{3}\). (b) Identify the conjugate acids of the given Bronsted-Lowry bases:
03

(i) Conjugate acid of \(\mathrm{O}^{2-}\)

To find the conjugate acid of \(\mathrm{O}^{2-}\), we need to add a proton (H鈦�) to its structure to form the conjugate acid: \[ \mathrm{O}^{2-} + \mathrm{H}^+ \rightarrow \mathrm{OH}^{-}\] So the conjugate acid of \(\mathrm{O}^{2-}\) is \(\mathrm{OH}^{-}\).
04

(ii) Conjugate acid of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

To find the conjugate acid of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\), we need to add a proton (H鈦�) to its structure to form the conjugate acid: \[ \mathrm{H}_{2} \mathrm{PO}_{4}^{-} + \mathrm{H}^+ \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}\] So the conjugate acid of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is \(\mathrm{H}_{3} \mathrm{PO}_{4}\).

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Key Concepts

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

Conjugate Base
In the Bronsted-Lowry theory of acids and bases, a conjugate base is what is left after an acid has donated a proton (H鈦�) during a chemical reaction. Understanding this concept is important because it shows how substances can react and transform during chemical processes. For example, when the acid the acid \(\mathrm{HIO}_3\) loses a proton, it becomes the conjugate base, \(\mathrm{IO}_3^-\). Similarly, when the acid \(\mathrm{NH}_4^+\) donates a proton, it turns into its conjugate base, \(\mathrm{NH}_3\). In both cases, the initially acidic molecules release a proton, leaving behind the "base" forms.
  • Conjugate bases are formed by the removal of a proton from an acid.
  • The acid becomes a conjugate base when it donates that proton.
  • Knowing the conjugate base helps predict the behavior of the acid in reaction.
Conjugate Acid
Conjugate acids are formed when a Bronsted-Lowry base accepts a proton. This addition results in an increase in the positive charge of the original base, transforming it into its conjugate acid. For instance, when the base \(\mathrm{O}^{2-}\) receives a proton, it becomes the conjugate acid \(\mathrm{OH}^-\). Similarly, when \(\mathrm{H}_2\mathrm{PO}_4^-\) accepts a proton, it transforms into \(\mathrm{H}_3\mathrm{PO}_4\). Understanding this conversion is vital for predicting how substances can buffer or resist changes in pH.
  • Conjugate acids are formed by adding a proton to a base.
  • The base becomes a conjugate acid when it accepts the proton.
  • This knowledge helps in determining the potential direction of chemical reactions.
Proton Transfer
Proton transfer is a fundamental chemical process in which protons (H鈦� ions) are transferred from one molecule to another. This movement is the basis for reactions between acids and bases according to the Bronsted-Lowry theory. During these reactions, a proton is transferred from the acid, turning it into its conjugate base, while the base accepting the proton turns into its conjugate acid. For example, in the reaction of \(\mathrm{HIO}_3\) with \(\mathrm{O}^{2-}\), the \(\mathrm{O}^{2-}\) accepts a proton from \(\mathrm{HIO}_3\), demonstrating proton transfer and forming their respective conjugate pairs.
  • Proton transfer is a key mechanism in acid-base reactions.
  • This process results in the formation of conjugate acid-base pairs.
  • Understanding proton transfer is essential for grasping chemical reaction dynamics.
Chemical Reactions
Chemical reactions entail the transformation of substances through bond making and breaking, often involving electron transfer or proton exchange. In the context of Bronsted-Lowry acids and bases, chemical reactions often focus on proton exchange. This involves acids donating protons and bases accepting protons, resulting in new substances known as conjugate acids and bases. For example, through chemical reactions \(\mathrm{NH}_4^+ + \mathrm{OH}^-\) can form water and \(\mathrm{NH}_3\) by way of proton exchange.
  • Chemical reactions involve the transformation of molecules into new products.
  • In acid-base reactions, the main exchange is protons between acids and bases.
  • Understanding chemical reactions allows for the prediction and control of chemical behavior.

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Most popular questions from this chapter

Phenylacetic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\right)\) is one of the substances that accumulates in the blood of people with phenylketonuria, an inherited disorder that can cause mental retardation or even death. A \(0.085 \mathrm{M}\) solution of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\) has a pH of \(2.68 .\) Calculate the \(K_{a}\) value for this acid.

Although the acid-dissociation constant for phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) is listed in Appendix \(\mathrm{D}\), the base - \(\mathrm{d}\) issociation constant for the phenolate ion \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}\right)\) is not. (a) Explain why it is not necessary to list both \(K_{a}\) for phenol and \(K_{b}\) for the phenolate ion. (b) Calculate \(K_{b}\) for the phenolate ion. (c) Is the phenolate ion a weaker or stronger base than ammoria?

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons\) (c) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\)

The iodate ion is reduced by sulfite according to the following reaction: $$ \mathrm{IO}_{3}^{-}(a q)+3 \mathrm{SO}_{3}{ }^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+3 \mathrm{SO}_{4}{ }^{2-}(a q) $$ The rate of this reaction is found to be first order in \(\mathrm{IO}_{3}\). first order in \(\mathrm{SO}_{3}{ }^{2-}\), and first order in \(\mathrm{H}^{+}\). (a) Write the rate law for the reaction. (b) By what factor will the rate of the reaction change if the \(\mathrm{pH}\) is lowered from \(5.00\) to 3.50? Does the reaction proceed faster or slower at the lower pH? (c) By using the concepts discussed in Section 14.6, explain how the reaction can be pH- dependent

The volume of an adult's stomach ranges from about \(50 \mathrm{~mL}\) when empty to \(1 \mathrm{~L}\) when full. If the stomach volume is \(400 \mathrm{~mL}\) and its contents have a \(\mathrm{pH}\) of 2 , how many moles of \(\mathrm{H}^{+}\) does the stomach contain? Assuming that all the \(\mathrm{H}^{+}\) comes from \(\mathrm{HCl}\), how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?
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