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Chapter 16: Problem 101

The odor of fish is due primarily to amines, especially methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right) .\) Fish is often served with a wedge of lemon, which contains citric acid. The amine and the acid react forming a product with no odor, thereby making the less-than-fresh fish more appetizing. Using data from Appendix \(\mathrm{D},\) calculate the equilibrium constant for the reaction of citric acid with methylamine, if only the first proton of the citric acid \(\left(K_{a 1}\right)\) is important in the neutralization reaction.

Short Answer

Expert verified
The equilibrium constant for the reaction of citric acid with methylamine, considering only the first proton of the citric acid \((K_{a1})\), can be calculated using the relation: \[ K_{eq} = \frac{K_{a1} K_{b}}{K_w} \] Using the given values of \(K_{a1}\) and \(K_{b}\) from Appendix D, and considering \(K_w = 1.0 \times 10^{-14}\) at \(25°C\), we can calculate \(K_{eq}\) for this reaction.

Step by step solution

01

Write the chemical equation for the reaction of citric acid with methylamine under equilibrium conditions.

The chemical reaction between citric acid \((H_3A)\) and methylamine \((CH_3NH_2)\) can be represented, considering only the first proton, as: \[ H_{3}A + CH_{3} NH_{2} \rightleftharpoons H_{2}A^{-} + CH_{3} NH_{3}^{+} \]
02

Write the expression for the equilibrium constant for the reaction.

We can write the expression for the equilibrium constant \((K_{eq})\) for the given reaction as follows: \[ K_{eq} = \frac{[H_{2}A^{-}][CH_{3} NH_{3}^{+}]}{[H_{3}A][CH_{3} NH_{2}]} \]
03

Relate \(K_{eq}\) to the ionization constant, \(K_{a}\) of citric acid.

In order to find \(K_{eq}\), we need to relate it to the ionization constant of citric acid for the first proton, also known as \(K_{a1}\). When the first proton is removed from the citric acid molecule \((H_3A)\), we have the following equilibrium: \[ H_{3}A \rightleftharpoons H_{2}A^{-} + H^{+} \] \[ K_{a1} = \frac{[H_{2}A^{-}][H^{+}]}{[H_{3}A]} \]
04

Relate \(K_{eq}\) to the basicity constant, \(K_{b}\) of methylamine.

Similarly, we need to relate \(K_{eq}\) to the basicity constant of methylamine, \(K_{b}\). We can represent the base ionization constant of methylamine with the following equilibrium: \[ CH_{3} NH_{2} + H_{2}O \rightleftharpoons CH_{3} NH_{3}^{+} + OH^{-} \] \[ K_{b} = \frac{[CH_{3} NH_{3}^{+}][OH^{-}]}{[CH_{3} NH_{2}]} \]
05

Calculate \(K_{eq}\) using \(K_{a1}\) and \(K_{b}\).

We know that for a reaction between an acid and a base, the equilibrium constant \(K_{eq}\) is related to \(K_a\) and \(K_b\) as follows: \[ K_{eq} = \frac{K_{a1} K_{b}}{K_w} \] where \(K_w = [H^{+}][OH^{-}]\) is the ion product of water and is equal to \(1.0 \times 10^{-14}\) at \(25°C\). Now we can use the given values of \(K_{a1}\) and \(K_{b}\) from Appendix D, and use the equation above to calculate \(K_{eq}\) for the reaction between citric acid and methylamine at equilibrium.

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Most popular questions from this chapter

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?

codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a weak organic base. \(\mathrm{A} 5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a pH of \(9.95 .\) Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{pK}_{b}\) for this base?

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{BrO}^{-}\) or \(\mathrm{ClO}^{-},(\mathbf{b}) \mathrm{BrO}^{-}\) or \(\mathrm{BrO}_{2}^{-}\) (c) \(\mathrm{HPO}_{4}^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}.\)

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

If a neutral solution of water, with \(\mathrm{pH}=7.00\) , is cooled to \(10^{\circ} \mathrm{C},\) the ph rises to \(7.27 .\) Which of the following three statements is correct for the cooled water: (i) \(\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right],\) (ii) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right],\) or (iii) \(\left[\mathrm{H}^{+}\right]<\left[\mathrm{OH}^{-}\right] ?\)
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