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Chapter 14: Problem 1

Consider two beakers of pure water at different temperatures. How do their \(\mathrm{pH}\) values compare? Which is more acidic? more basic? Explain.

Short Answer

Expert verified
In conclusion, when comparing two beakers of pure water at different temperatures, the beaker with a higher temperature will have a lower pH value, making it relatively more acidic. Conversely, the beaker with a lower temperature will have a higher pH value, making it relatively more basic. This is due to the temperature-dependent nature of the ion product of water and the increased dissociation of water molecules at higher temperatures. However, both beakers are still considered neutral, as their H+ and OH- ion concentrations are equal.

Step by step solution

01

Understand the dissociation of water molecules and pH

Water molecules can dissociate into hydrogen ions (H+) and hydroxide ions (OH-), a process called autoionization. The equilibrium constant for this reaction, called the ion product of water (Kw), is defined as: Kw = [H+][OH-] The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration: pH = -log([H+]) In pure water at standard conditions (25°C, 1 atm), the concentration of both ions is equal, and Kw=1.0×10^(-14).
02

Determine the effect of temperature on the ion product of water

The ion product of water (Kw) is temperature-dependent. As the temperature of water increases, the equilibrium constant for the autoionization of water also increases, causing more dissociation of water molecules and higher concentrations of H+ and OH- ions in the solution. This means that the pH of pure water decreases with increasing temperature, as pH depends on the concentration of H+ ions.
03

Compare the pH values and determine the acidic and basic nature of the water samples

Since the pH of pure water decreases with increasing temperature, a beaker of pure water at a higher temperature will have a lower pH value (more acidic conditions) compared to a beaker of pure water at a lower temperature (more basic conditions). However, it's important to note that pure water is still considered neutral (neither acidic nor basic) in both cases, as the concentrations of H+ and OH- ions are equal.
04

Conclusion

In conclusion, the pH values of two beakers of pure water at different temperatures will differ due to the temperature-dependent nature of the ion product of water. As the temperature increases, the pH value of pure water will decrease, causing it to be more acidic compared to pure water at a lower temperature, which will be more basic. Nonetheless, pure water remains neutral in both cases, as the concentrations of H+ and OH- ions are equal.

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Most popular questions from this chapter

The \(\mathrm{pOH}\) of a sample of baking soda dissolved in water is \(5.74\) at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH},\left[\mathrm{H}^{+}\right]\), and \(\left[\mathrm{OH}^{-}\right]\) for this sample. Is the solution acidic or basic?

For the reaction of hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) in water. $$ \mathrm{H}_{2} \mathrm{NNH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{NNH}_{3}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{b}}\) is \(3.0 \times 10^{-6}\). Calculate the concentrations of all species and the \(\mathrm{pH}\) of a \(2.0 \mathrm{M}\) solution of hydrazine in water.

Acrylic acid \(\left(\mathrm{CH}_{2}=\mathrm{CHCO}_{2} \mathrm{H}\right)\) is a precursor for many important plastics. \(K_{\mathrm{a}}\) for acrylic acid is \(5.6 \times 10^{-5}\). a. Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of acrylic acid. b. Calculate the percent dissociation of a \(0.10 \mathrm{M}\) solution of acrylic acid. c. Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of sodium acrylate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.4 \times 10^{-5}\).

A sample containing \(0.0500 \mathrm{~mol} \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is dissolved in enough water to make \(1.00 \mathrm{~L}\) of solution. This solution contains hydrated \(\mathrm{SO}_{4}{ }^{2-}\) and \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\) ions. The latter behaves as an acid: $$\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}(a q)+\mathrm{H}^{+}(a q)$$ a. Calculate the expected osmotic pressure of this solution at \(25^{\circ} \mathrm{C}\) if the above dissociation is negligible. b. The actual osmotic pressure of the solution is \(6.73\) atm at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{a}}\) for the dissociation reaction of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\). (To do this calculation, you must assume that none of the ions goes through the semipermeable membrane. Actually, this is not a great assumption for the tiny \(\mathrm{H}^{+}\) ion.)
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