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Chapter 16: Problem 2

Write the formula of the conjugate base in the reaction of each acid with water. (a) \(\mathrm{HIO}_{3} ;\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (c) \(\mathrm{HPO}_{4}^{2-} ;\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)

Short Answer

Expert verified
The formulas for the conjugate bases are (a) \( IO_{3}^{-} \); (b) \( C_{6}H_{5}COO^{-} \); (c) \( PO_{4}^{3-} \); (d) \( C_{2}H_{5}NH_{2} \)

Step by step solution

01

Write the Acid Reaction with Water

For each given acid, write an equation of it reacting with water. For example, the reaction of \(HIO_{3}\) with water can be written as: \(HIO_{3} + H_{2}O ↔ H_{3}O^{+} + IO_{3}^{-}\)
02

Identify the Conjugate Base

The conjugate base of the acid is the substance that remains after the acid donates a proton (H+). This will be obvious from the right side of the equation. For our example in step 1, \(IO_{3}^{-}\) is the conjugate base of the acid \(HIO_{3}\)
03

Repeat for the remaining acids

Repeat the above steps for each given acid: \(C_{6}H_{5}COOH\), \(HPO_{4}^{2-}\), and \(C_{2}H_{5}NH_{3}^{+}\). The conjugate bases obtained will be \(C_{6}H_{5}COO^{-}\), \(PO_{4}^{3-}\), and \(C_{2}H_{5}NH_{2}\) respectively.
04

Validate the Results

For each equation outlined in the prior steps, ensure that the conjugate base makes sense. This means that it should have one less hydrogen atom and it is one step more negative (or less positive) than the original acid. Validate this for each conjugate base obtained.

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Most popular questions from this chapter

Here is a way to test the validity of the statement made on page 719 in conjunction with the three key ideas governing the ionization of polyprotic acids. Determine the \(\mathrm{pH}\) of \(0.100 \mathrm{M}\) succinic acid in two ways: first by assuming that \(\mathrm{H}_{3} \mathrm{O}^{+}\) is produced only in the first ionization step, and then by allowing for the possibility that some \(\mathrm{H}_{3} \mathrm{O}^{+}\) is also produced in the second ionization step. Compare the results, and discuss the significance of your finding. $$\begin{array}{c} \mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HC}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{-} \\ &K_{\mathrm{a}_{1}}=6.2 \times 10^{-5} \\ \mathrm{HC}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{2-} \\ &K_{\mathrm{a}_{2}}=2.3 \times 10^{-6} \end{array}$$

Explain the important distinctions between each pair of terms: (a) Bronsted- Lowry acid and base; (b) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\mathrm{pH} ;\) (c) \(K_{\mathrm{a}}\) for \(\mathrm{NH}_{4}^{+}\) and \(K_{\mathrm{b}}\) for \(\mathrm{NH}_{3} ;\) (d) leveling effect and electron- withdrawing effect.

Use Lewis structures to diagram the following reaction in the manner of reaction (16.20) $$\mathrm{H}_{2} \mathrm{O}+\mathrm{SO}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{3}$$ Identify the Lewis acid and Lewis base.

What mass of benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), would you dissolve in \(350.0 \mathrm{mL}\) of water to produce a solution with a \(\mathrm{pH}=2.85 ?\) $$\begin{aligned} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} & \\ K_{\mathrm{a}}=6.3 \times 10^{-5} \end{aligned}$$

Explain why trichloroacetic acid, \(\mathrm{CCl}_{3} \mathrm{COOH},\) is a stronger acid than acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\).
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