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, \(\mathrm{HPO}_{4}^{2-} ;(\mathbf{c})\) benzoate ion, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}$

Short Answer

Expert verified
(a) Chemical equation for propylamine with water: \[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}\] \(K_{b}\) expression for propylamine: \[K_{b} = \frac{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{2}]} \] (b) Chemical equation for monohydrogen phosphate ion with water: \[\mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{OH}^{-}\] \(K_{b}\) expression for monohydrogen phosphate ion: \[K_{b} = \frac{[\mathrm{H}_{2}\mathrm{PO}_{4}^{-}][\mathrm{OH}^{-}]}{[\mathrm{HPO}_{4}^{2-}]} \] (c) Chemical equation for benzoate ion with water: \[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H} + \mathrm{OH}^{-}\] \(K_{b}\) expression for benzoate ion: \[K_{b} = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}]} \]

Step by step solution

01

(a) Chemical equation for propylamine with water

Propylamine, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2}\), is a weak base, which means it will accept a proton from water. The chemical equation for this reaction is: \[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}\]
02

(a) \(K_{b}\) expression for propylamine

Now, we can write the base ionization constant (\(K_{b}\)) expression for propylamine: \[K_{b} = \frac{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{2}]} \]
03

(b) Chemical equation for monohydrogen phosphate ion with water

Monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-}\), is an amphoteric ion (can act as both acid and base). In this case, we will consider it as a weak base that accepts a proton from water: \[\mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{OH}^{-}\]
04

(b) \(K_{b}\) expression for monohydrogen phosphate ion

The base ionization constant (\(K_{b}\)) expression for monohydrogen phosphate ion is: \[K_{b} = \frac{[\mathrm{H}_{2}\mathrm{PO}_{4}^{-}][\mathrm{OH}^{-}]}{[\mathrm{HPO}_{4}^{2-}]} \]
05

(c) Chemical equation for benzoate ion with water

Benzoate ion, \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}\), acts as a weak base and will accept a proton from water: \[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H} + \mathrm{OH}^{-}\]
06

(c) \(K_{b}\) expression for benzoate ion

Lastly, we can write the base ionization constant (\(K_{b}\)) expression for the benzoate ion: \[K_{b} = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}]} \]

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Most popular questions from this chapter

Write the chemical equation and the \(K_{a}\) expression for the dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: \((\mathbf{a}) \mathrm{HNO}_{2},\) (b) \(\mathrm{ClH}_{2} \mathrm{CCOOH}\).

Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but they are frequently much more soluble as their acid salts. Assuming that \(\mathrm{pH}\) in the stomach is 2.5 , indicate whether each of the following compounds would be present in the stomach as the neutral base or in the protonated form: nicotine, $K_{b}=7 \times 10^{-7}\(; caffeine, \)K_{b}=4 \times 10^{-14} ;\( strychnine, \)K_{b}=1 \times 10^{-6} ;\( quinine, \)K_{b}=1.1 \times 10^{-6} .$

Calculate the number of \(\mathrm{H}^{+}(a q)\) ions in \(1.0 \mathrm{~mL}\) of pure water at \(25^{\circ} \mathrm{C}\).

By what factor does \(\left[\mathrm{H}^{+}\right]\) change for a pH change of (a) 3.0 units, (b) 0.3 units?

Succinic acid $\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{4}\right),\( which we will denote \)\mathrm{H}_{2} \mathrm{Suc}$ is a biologically relevant diprotic acid with the structure shown below. At \(25^{\circ} \mathrm{C}\), the acid-dissociation constants for succinic acid are \(K_{a 1}=6.9 \times 10^{-5}\) and \(K_{a 2}=2.5 \times 10^{-6} .\) (a) Determine the pH of a \(0.32 \mathrm{M}\) solution of $\mathrm{H}_{2} \mathrm{Suc}\( at \)25^{\circ} \mathrm{C}$, assuming that only the first dissociation is relevant. (b) Determine the molar concentration of \(\mathrm{Suc}^{2-}\) in the solution in part (a). (c) Is the assumption you made in part (a) justified by the result from part (b)? (d) Will a solution of the salt NaHSuc be acidic, neutral, or basic?

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