Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} ;(\mathbf{b})\) carbonate ion, \(\mathrm{CO}_{3}^{2-} ;(\mathbf{c})\) formate ion, \(\mathrm{CHO}_{2}^{-} .\)

Short Answer

Expert verified
(a) Dimethylamine reacts with water: \((\mathrm{CH}_{3})_{2} \mathrm{NH} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow (\mathrm{CH}_{3})_{2} \mathrm{NH}_{2}^{+} + \mathrm{OH}^{-}\), with \(K_{b} = \frac{[ (\mathrm{CH}_{3})_{2} \mathrm{NH}_{2}^{+}] [\mathrm{OH}^{-}]}{[(\mathrm{CH}_{3})_{2} \mathrm{NH}]}\) (b) Carbonate ion reacts with water: \(\mathrm{CO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-}\), with \(K_{b} = \frac{[\mathrm{HCO}_{3}^{-}] [\mathrm{OH}^{-}]}{[\mathrm{CO}_{3}^{2-}]}\) (c) Formate ion reacts with water: \(\mathrm{CHO}_{2}^{-} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{CHO}_{2}\mathrm{H} + \mathrm{OH}^{-}\), with \(K_{b} = \frac{[\mathrm{CHO}_{2}\mathrm{H}] [\mathrm{OH}^{-}]}{[\mathrm{CHO}_{2}^{-}]}\)

Step by step solution

01

(a) Reaction of dimethylamine with water

Dimethylamine reacts with water to form dimethylammonium ion and hydroxide ion. The chemical equation of this reaction is: \((\mathrm{CH}_{3})_{2} \mathrm{NH} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow (\mathrm{CH}_{3})_{2} \mathrm{NH}_{2}^{+} + \mathrm{OH}^{-}\)
02

(a) \(K_{b}\) expression for the reaction of dimethylamine with water

Based on the chemical equation, the \(K_{b}\) expression can be written as: \[K_{b} = \frac{[ (\mathrm{CH}_{3})_{2} \mathrm{NH}_{2}^{+}] [\mathrm{OH}^{-}]}{[(\mathrm{CH}_{3})_{2} \mathrm{NH}]}\]
03

(b) Reaction of carbonate ion with water

Carbonate ion reacts with water to form bicarbonate ion and hydroxide ion. The chemical equation of this reaction is: \(\mathrm{CO}_{3}^{2-} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-}\)
04

(b) \(K_{b}\) expression for the reaction of carbonate ion with water

Based on the chemical equation, the \(K_{b}\) expression can be written as: \[K_{b} = \frac{[\mathrm{HCO}_{3}^{-}] [\mathrm{OH}^{-}]}{[\mathrm{CO}_{3}^{2-}]}\]
05

(c) Reaction of formate ion with water

Formate ion reacts with water to form formic acid and hydroxide ion. The chemical equation of this reaction is: \(\mathrm{CHO}_{2}^{-} + \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{CHO}_{2}\mathrm{H} + \mathrm{OH}^{-}\)
06

(c) \(K_{b}\) expression for the reaction of formate ion with water

Based on the chemical equation, the \(K_{b}\) expression can be written as: \[K_{b} = \frac{[\mathrm{CHO}_{2}\mathrm{H}] [\mathrm{OH}^{-}]}{[\mathrm{CHO}_{2}^{-}]}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) are both derivatives of benzene. Benzoic acid is an acid with \(K_{a}=6.3 \times 10^{-5}\) and aniline is a base with \(K_{a}=4.3 \times 10^{-10} .\) (a) What are the conjugate base of benzoic acid and the conjugate acid of aniline? (b) Anilinium chloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\right)\) is a strong electrolyte that dissociates into anilinium ions \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right)\) and chloride ions. Which will be more acidic, a 0.10\(M\) solution of benzoic acid or a 0.10 M solution of anilinium chloride? (c) What is the value of the equilibrium constant for the following equilibrium? $$\begin{array}{c}{\mathrm{C}_{6} \mathrm{H}_{5} \operatorname{COOH}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(a q) \rightleftharpoons} \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad {\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(a q)}\end{array}$$

The volume of an adult's stomach ranges from about 50 mL when empty to 1 when full. If the stomach volume is 400 mL and its contents have a 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?

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{OH}^{-}\right]=0.00045 M ;\) (b) \(\left[\mathrm{OH}^{-}\right]=8.8 \times 10^{-9} \mathrm{M} ;\) (c) a solution in which \(\left[\mathrm{OH}^{-}\right]\) is 100 times greater than \(\left[\mathrm{H}^{+}\right]\) .

The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} .\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature.

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right),\) a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, containing 500 \(\mathrm{mg}\) of acetylsalicylic acid each, in 250 \(\mathrm{mL}\) of water?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free