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Chapter 17: Problem 73

The neutralization of \(\mathrm{NaOH}\) by \(\mathrm{HCl}\) is represented in equation (1), and the neutralization of \(\mathrm{NH}_{3}\) by HCl in equation (2). 1. \(\mathrm{OH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O} \quad K=?\) 2\. \(\mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \quad K=?\) (a) Determine the equilibrium constant \(K\) for each reaction. (b) Explain why each neutralization reaction can be considered to go to completion.

Short Answer

Expert verified
The equilibrium constant K for the first reaction can be represented as \(1/[OH-][H3O+]\) and for the second reaction it can be represented as \([NH4+] / [NH3]\). Each neutralization reaction can be considered to go to completion as these reactions typically proceed until one of the reactants is exhausted due to the high reactivity of both acids and bases.

Step by step solution

01

Calculate equilibrium constant K for the first reaction

For the first reaction, \(\mathrm{OH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \to 2 \mathrm{H}_{2} \mathrm{O}\), the equilibrium constant \(K\) can be determined by setting it equal to the concentrations of the products divided by the concentrations of the reactants. However, since water is a product and it remains fairly constant, the equilibrium constant is primarily determined by the concentration of the reactants, which is \([OH-][H3O+]\). Therefore, for this reaction, \(K = [H2O] / [OH-][H3O+] = 1/[OH-][H3O+]\).
02

Calculate equilibrium constant K for the second reaction

For the second reaction, \(\mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \to \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O}\), the equilibrium constant \(K\) can be determined in a similar manner. Since the concentration of water remains fairly constant, the equilibrium constant is primarily determined by the concentration of the reactants, which is \([NH3][H3O+]\). Therefore, for this reaction, \(K = [NH4+][H2O] / [NH3][H3O+] = [NH4+] / [NH3]\).
03

Explain why each neutralization reaction can be considered to go to completion

An acid-base neutralization reaction can be considered to go to completion because it is a process where an acid and a base react quantitatively with each other. In such reactions, the acid and base become less acidic and less basic, respectively, as they react with each other, leading to the formation of water and a type of salt. This reaction proceeds until one of the reactants is exhausted. Due to the high reactivity of both acids and bases, these reactions typically proceed to completion, meaning that there are no leftover reactants.

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Most popular questions from this chapter

Explain the important distinctions between each pair of terms: (a) buffer capacity and buffer range; (b) hydrolysis and neutralization; (c) first and second equivalence points in the titration of a weak diprotic acid; (d) equivalence point of a titration and end point of an indicator.

The pH of a solution of \(19.5 \mathrm{g}\) of malonic acid in \(0.250 \mathrm{L}\) is \(1.47 .\) The pH of a \(0.300 \mathrm{M}\) solution of sodium hydrogen malonate is 4.26. What are the values of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for malonic acid?

In 1922 Donald D. van Slyke ( J. Biol. Chem., 52, 525) defined a quantity known as the buffer index: \(\beta=\mathrm{d} C_{\mathrm{b}} / \mathrm{d}(\mathrm{pH}),\) where \(\mathrm{d} C_{\mathrm{b}}\) represents the increment of moles of strong base to one liter of the buffer. For the addition of a strong acid, he wrote \(\beta=-\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH})\) By applying this idea to a monoprotic acid and its conjugate base, we can derive the following expression: \(\beta=2.303\left(\frac{K_{w}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\frac{\mathrm{CK}_{\mathrm{a}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left(\mathrm{K}_{\mathrm{a}}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\right)^{2}}\right)\) where \(C\) is the total concentration of monoprotic acid and conjugate base. (a) Use the above expression to calculate the buffer index for the acetic acid buffer with a total acetic acid and acetate ion concentration of \(2.0 \times 10^{-2}\) and a \(\mathrm{pH}=5.0\) (b) Use the buffer index from part (a) and calculate the \(\mathrm{pH}\) of the buffer after the addition of of a strong acid. (Hint: Let \(\left.\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH}) \approx \Delta C_{\mathrm{a}} / \Delta \mathrm{pH} .\right)\) (c) Make a plot of \(\beta\) versus \(\mathrm{pH}\) for a \(0.1 \mathrm{M}\) acetic acid buffer system. Locate the maximum buffer index as well as the minimum buffer indices.

Both sodium hydrogen carbonate (sodium bicarbonate) and sodium hydroxide can be used to neutralize acid spills. What is the pH of \(1.00 \mathrm{M} \mathrm{NaHCO}_{3}(\mathrm{aq})\) and of \(1.00 \mathrm{M} \mathrm{NaOH}(\mathrm{aq}) ?\) On a per-liter basis, do these two solutions have an equal capacity to neutralize acids? Explain. On a per-gram basis, do the two solids, \(\mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{NaOH}(\mathrm{s}),\) have an equal capacity to neutralize acids? Explain. Why do you suppose that \(\mathrm{NaHCO}_{3}\) is often preferred to \(\mathrm{NaOH}\) in neutralizing acid spills?

The Henderson-Hasselbalch equation can be written as \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}-\log \left(\frac{1}{\alpha}-1\right)\) where \(\alpha=\frac{\left[\mathrm{A}^{-}\right]}{\left[\mathrm{A}^{-}\right]+[\mathrm{HA}]}\) Thus, the degree of ionization \((\alpha)\) of an acid can be determined if both the \(\mathrm{pH}\) of the solution and the \(\mathrm{p} K_{\mathrm{a}}\) of the acid are known. (a) Use this equation to plot the pH versus the degree of ionization for the second ionization constant of phosphoric acid \(\left(K_{\mathrm{a}}=6.3 \times 10^{-8}\right)\) (b) If \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) what is the degree of ionization? (c) If the solution had a pH of 6.0 what would the value of \(\alpha\) be?
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