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Chapter 16: Problem 72

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

Expert verified
(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, HPO4虏鈦�, 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鈦�, 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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Key Concepts

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

Understanding Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry, where reactions can proceed in both the forward and reverse directions until the rates of the forward and reverse reactions are equal. At this point, the concentrations of the reactants and products remain constant. This does not mean the amounts are equal, just that they are unchanging over time.
The implication of equilibrium in a chemical reaction can be seen in the equations for weak bases reacting with water we鈥檝e discussed. These equations have a double arrow (鈫�), signifying that equilibrium is reached between the compounds on either side.
  • Equilibrium acknowledges that the forward reaction, where the base reacts with water, is occurring at the same rate as the reverse, where the products recombine to form the reactants again.
  • Understanding this dynamic helps in comprehending the importance of equilibrium constants such as the base dissociation constant, denoted as \(K_b\), which quantifies the extent to which a weak base reacts with water to form its conjugate acid and hydroxide ion.
In practice, most reactions involving weak bases and their equilibria require knowing the concentrations of involved species to calculate equilibrium concentrations using \(K_b\).
Exploring Weak Bases
Weak bases are substances that partially ionize in water. Unlike strong bases which completely dissociate, weak bases like propylamine, monohydrogen phosphate, and benzoate ion only dissociate to a small extent. This partial ionization is pivotal to their behavior in chemical equilibrium.When a weak base
  • reacts with water, it accepts a proton (\(H^+\)) from a water molecule, forming its conjugate acid and a hydroxide ion \( (\text{OH}^- ) \).
  • This process is captured by equations that include equilibrium conditions since the reaction doesn't go to completion.
For example, propylamine reacts with water to form propylammonium ion and hydroxide ion:
\[ \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{3}\mathrm{H}_{7}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-} \]Being weak means there is an established equilibrium where the base is only partially converted into ions, governed by the \(K_b\) expression, representing the concentrations of these species in equilibrium.
Understanding Conjugate Acid-Base Pairs
Conjugate acid-base pairs help explain the behavior of acids and bases through the transfer of protons. When a base gains a proton, it becomes a conjugate acid, and when an acid loses a proton, it becomes a conjugate base. This concept is crucial in understanding reactions involving weak bases.
In the reactions we considered:
  • Propylamine鈥檚 conjugate acid is propylammonium ion \( (\mathrm{C}_{3} \mathrm{H}_{7}\mathrm{NH}_{3}^{+}) \).
  • Monohydrogen phosphate becomes dihydrogen phosphate \( (\mathrm{H}_{2}\mathrm{PO}_{4}^{-}) \) upon protonation.
  • Benzoate ion forms benzoic acid \( (\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}) \) as its conjugate acid.
This relationship is central in equilibrium considerations as the extent to which a base forms its conjugate acid and hydroxide ion directly impacts the calculations involving the \(K_b\) constant. Conjugate pairs showcase the reversible nature of these reactions, reinforcing the concept that adding or removing a component can shift equilibrium, illustrating Le Ch芒telier鈥檚 principle.

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Most popular questions from this chapter

Triethylamine, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N},\) has a \(\mathrm{p} K_{b}\) value of \(2.99 .\) Is triethylamine a stronger base than ammonia, \(\mathrm{NH}_{3} ?\)

Label each of the following as being a strong acid, a weak acid, or a species with negligible acidity. In each case write the formula of its conjugate base, and indicate whether the conjugate base is a strong base, a weak base, or a species with negligible basicity: (a) \(\mathrm{HCOOH},\) (b) \(\mathrm{H}_{2},\) (c) \(\mathrm{CH}_{4}\), (d) \(\mathrm{HF}\) (e) \(\mathrm{NH}_{4}^{+}\).

Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base of each on the right side: $$ \text { (a) } \mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q) $$ (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) $$ \mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q) $$

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(8.5 \times 10^{-3} \mathrm{MHBr}\), (b) \(1.52 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) in \(575 \mathrm{~mL}\) of solution, (c) \(5.00 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HClO}_{4}\) diluted to \(50.0 \mathrm{~mL}\), (d) a solution formed by mixing \(10.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HBr}\) with \(20.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\)

(a) Given that \(K_{a}\) for acetic acid is \(1.8 \times 10^{-5}\) and that for hypochlorous acid is \(3.0 \times 10^{-8}\), which is the stronger acid? (b) Which is the stronger base, the acetate ion or the hypochlorite ion? (c) Calculate \(K_{b}\) values for \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) and \(\mathrm{ClO}^{-}\).
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