Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) propylamine,\(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} ;\) (b) monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-}\); (c) benzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\).

Short Answer

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(a) The chemical equation for the reaction of propylamine with water is: \( \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} (\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O} (\mathrm{l}) \rightleftharpoons \mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+}(\mathrm{aq}) + \mathrm{OH}^{-} (\mathrm{aq}) \) The \(K_{b}\) expression is: \( K_{b} = \frac{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{2}]} \) (b) The chemical equation for the reaction of monohydrogen phosphate ion with water is: \( \mathrm{HPO}_{4}^{2-} (\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O} (\mathrm{l}) \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^{-} (\mathrm{aq}) + \mathrm{OH}^{-} (\mathrm{aq}) \) The \(K_{b}\) expression is: \( K_{b} = \frac{[\mathrm{H}_{2}\mathrm{PO}_{4}^{-}][\mathrm{OH}^{-}]}{[\mathrm{HPO}_{4}^{2-}]} \) (c) The chemical equation for the reaction of benzoate ion with water is: \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-} (\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O} (\mathrm{l}) \rightleftharpoons\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH} (\mathrm{aq}) + \mathrm{OH}^{-} (\mathrm{aq}) \) The \(K_{b}\) expression is: \( K_{b} = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}]} \)

Step by step solution

01

Chemical equation and Kb expression for propylamine with water

Propylamine, C3H7NH2, is a weak base that reacts with water to form the conjugate acid and hydroxide ion. The chemical equation can be written as: \( \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} (\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O} (\mathrm{l}) \rightleftharpoons \mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+}(\mathrm{aq}) + \mathrm{OH}^{-} (\mathrm{aq}) \) The Kb expression for this reaction is: \( K_{b} = \frac{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{2}]} \)
02

Chemical equation and Kb expression for monohydrogen phosphate ion with water

Monohydrogen phosphate ion, HPO4²⁻, is also a weak base that reacts with water to form the conjugate acid and hydroxide ion. The chemical equation can be written as: \( \mathrm{HPO}_{4}^{2-} (\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O} (\mathrm{l}) \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^{-} (\mathrm{aq}) + \mathrm{OH}^{-} (\mathrm{aq}) \) The Kb expression for this reaction is: \( K_{b} = \frac{[\mathrm{H}_{2}\mathrm{PO}_{4}^{-}][\mathrm{OH}^{-}]}{[\mathrm{HPO}_{4}^{2-}]} \)
03

Chemical equation and Kb expression for benzoate ion with water

Benzoate ion, C6H5COO⁻, reacts with water to form the conjugate acid and hydroxide ion. The chemical equation can be written as: \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-} (\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O} (\mathrm{l}) \rightleftharpoons\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH} (\mathrm{aq}) + \mathrm{OH}^{-} (\mathrm{aq}) \) The Kb expression for this reaction is: \( K_{b} = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}]} \)

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Most popular questions from this chapter

Calculate the percent ionization of hydrazoic acid \(\left(\mathrm{HN}_{3}\right)\) in solutions of each of the following concentrations \(\left(K_{a}\right.\) is given in Appendix D): (a) $0.400 \mathrm{M}\(, (b) \)0.100 \mathrm{M}\(, (c) \)0.0400 \mathrm{M}$.

(a) The hydrogen oxalate ion \(\left(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\right)\) is amphiprotic. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base toward water. (b) What is the conjugate acid of \(\mathrm{HC}_{2} \mathrm{O}_{4}\) ? What is its conjugate base?

Arrange the following \(0.10 \mathrm{M}\) solutions in order of increasing acidity (decreasing pH): (i) \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) (ii) \(\mathrm{NaNO}_{3}\), (iii) $$ \mathrm{CH}_{3} \mathrm{COONH}_{4} \text { , (iv) } \mathrm{NaF} \text { , (v) } \mathrm{CH}_{3} \mathrm{COONa} \text { . } $$

Calculate the percent ionization of propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)\) in solutions of each of the following concentrations \(\left(K_{a}\right.\) is given in Appendix \(\left.D\right):\) (a) \(0.250 \mathrm{M}\), (b) \(0.0800 \mathrm{M}\), (c) \(0.0200 \mathrm{M}\).

A \(0.100 M\) solution of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) is \(11.0 \%\) ionized. Using this information, calculate \(\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{H}^{+}\right],\) \(\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right],\) and \(K_{a}\) for chloroacetic acid.

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