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Chapter 16: Problem 120

At \(50^{\circ} \mathrm{C}\), the ion-product constant for $\mathrm{H}_{2} \mathrm{O}\( has the value \)K_{w}=5.48 \times 10^{-14} \cdot(\mathbf{a})$ What is the \(\mathrm{pH}\) of pure water at \(50^{\circ} \mathrm{C} ?\) (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

Short Answer

Expert verified
The pH of pure water at 50°C is 6.63, and based on the increase in the ion-product constant for water (Kw) with temperature, the enthalpy change (∆H) for the autoionization reaction of water is positive, indicating an endothermic reaction.

Step by step solution

01

Calculate the concentration of H3O+ ions

Be aware that in pure water, the concentration of H3O+ ions is equal to the concentration of OH- ions. Moreover, we know that Kw = [H3O+] * [OH-]. Since the concentrations are equal, we can rewrite the equation for Kw as follows: Kw = [H3O+]^2 We will use the given value for Kw at 50°C to find the concentration of H3O+ ions.
02

Solve for H3O+ ion concentration

We need to determine the [H3O+] from the given Kw at 50°C: \(K_w = 5.48 \times 10^{-14} = [H_3O^+]^2\) \( [H_3O^+] = \sqrt{5.48 \times 10^{-14}}\) \[ [H_3O^+] = 2.34 \times 10^{-7} \text{ M}\]
03

Calculate the pH of pure water

To find the pH of pure water, we use the relationship between pH and H3O+ ion concentration: \(pH = -\log ([H_3O^+])\) \(pH = -\log (2.34 \times 10^{-7})\) \[ pH = 6.63 \]
04

Analyze the change in Kw with temperature

The value of Kw increases as the temperature increases. At 25°C, the Kw value is \(1.0 \times 10^{-14}\), and at 50°C it is \(5.48 \times 10^{-14}\). This means that the autoionization reaction becomes more favorable at higher temperatures.
05

Predict the sign of ∆H for the autoionization reaction of water

Since the autoionization reaction of water becomes more favorable at higher temperatures, it indicates that the reaction is endothermic, absorbing heat from the surroundings. Therefore, the enthalpy change (∆H) for the autoionization reaction of water is positive. In summary, the pH of pure water at 50°C is 6.63, and the enthalpy change for the autoionization reaction of water is positive.

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Most popular questions from this chapter

Predict the stronger acid in each pair: (a) \(\mathrm{HCl}\) or HF; (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) or \(\mathrm{H}_{3} \mathrm{AsO}_{4} ;\) (c) \(\mathrm{HBrO}_{3}\) or \(\mathrm{HBrO}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) or \(\mathrm{HC}_{2} \mathrm{O}_{4} \overline{;} ;(\mathbf{e})\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) or phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right) .\)

(a) Give the conjugate base of the following Brønsted Lowry acids: (i) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-},\) (ii) HBr. (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{CN}^{-},\) (ii) \(\mathrm{HSO}_{4}^{-}\).

A particular sample of vinegar has a pH of 2.20 . If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\), calculate the concentration of acetic acid in the vinegar.

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 \(\left(K_{a}\right.\) isgiven in AppendixD): (a) \(0.250 M,(\mathbf{b}) 0.0800 M\) (c) \(0.0200 \mathrm{M}\)

Saccharin, a sugar substitute, is a weak acid with \(\mathrm{p} K_{a}=2.32\) at \(25^{\circ} \mathrm{C}\). It ionizes in aqueous solution as follows: $$ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) $$ What is the pH of a \(0.25 \mathrm{M}\) solution of this substance?
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