Chapter 16: Problem 18
Calculate the \(\mathrm{pH}\) of water at \(40^{\circ} \mathrm{C}\), given that \(K_{\mathrm{w}}\) is \(3.8 \times 10^{-14}\) at this temperature.
Short Answer
Expert verified
The pH of water at \(40^{\circ} \mathrm{C}\) is \(7.21\).
Step by step solution
01
Understand the Problem and Recognize the Formula
The \(K_{\mathrm{w}}\) indicates product of the molar concentration of H+ and OH- ions in water. For pure water, [H+] = [OH-]. So we can express the \(K_{\mathrm{w}}\) as \(K_{\mathrm{w}}\ =\ [H+]^2\). The task requires finding the pH, which is the negative logarithm of the [H+] concentration. Therefore it includes recognizing and using these formulas.
02
Solve for [H+]
First, solve the \(K_{\mathrm{w}}\ =\ [H+]^2\) for [H+]. This will require square rooting both sides of the equation. Thus, we get [H+] = \(\sqrt{K_{\mathrm{w}}}\), [H+] = \(\sqrt{3.8 \times 10^{-14}}\) and using the given value of \(K_{\mathrm{w}}\).
03
Calculate [H+]
Using the square root that we calculated in step 2, we calculate the [H+] concentration as: [H+] = \(6.1644 \times 10^{-8}\) M.
04
Calculate pH
Once we have the [H+], we can calculate the pH using the formula: pH = -log[H+]. Therefore, pH = -log(\(6.1644 \times 10^{-8}\)), and pH = \(7.21\) at \(40^{\circ} \mathrm{C}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kw (ion product of water)
The ion product of water, denoted as \( K_{\mathrm{w}} \), is a crucial concept when calculating pH and understanding solution chemistry. At any given temperature, \( K_{\mathrm{w}} \) is the equilibrium constant for the self-ionization of water, which can be represented by this chemical reaction:
Knowing \( K_{\mathrm{w}} \) allows us to determine the concentrations of \( \mathrm{H^+} \) and subsequently, the pH of the solution.
- \( \mathrm{H_2O} \leftrightharpoons \mathrm{H^+} + \mathrm{OH^-} \)
Knowing \( K_{\mathrm{w}} \) allows us to determine the concentrations of \( \mathrm{H^+} \) and subsequently, the pH of the solution.
temperature effect on pH
Temperature has a significant impact on pH levels, particularly in aqueous solutions like water. The pH of pure water is neutral and generally around 7.0 at 25°C. However, this neutrality changes with temperature shifts due to the temperature dependence of \( K_{\mathrm{w}} \).
At higher temperatures, self-ionization of water increases, leading to a higher \( K_{\mathrm{w}} \). For example, at 40°C, \( K_{\mathrm{w}} \) is \( 3.8 \times 10^{-14} \). This increase means that the concentration of \( \mathrm{H^+} \) ions is higher, resulting in a pH lower than 7.0 despite the solution remaining neutral.
At higher temperatures, self-ionization of water increases, leading to a higher \( K_{\mathrm{w}} \). For example, at 40°C, \( K_{\mathrm{w}} \) is \( 3.8 \times 10^{-14} \). This increase means that the concentration of \( \mathrm{H^+} \) ions is higher, resulting in a pH lower than 7.0 despite the solution remaining neutral.
- Greater \( K_{\mathrm{w}} \) values indicate more ionization at higher temperatures.
- Neutral pH decreases as temperature increases.
concentration of ions
In the context of water ionization, the concentration of ions like \( \mathrm{H^+} \) and \( \mathrm{OH^-} \) directly influences pH calculation. For pure water, these ion concentrations are equal, and their product is \( K_{\mathrm{w}} \). Solving for the \( \mathrm{H^+} \) concentration requires knowing \( K_{\mathrm{w}} \), using the equation:
Calculated concentrations of ions are crucial in determining the pH of solutions. For example, at a temperature of 40°C with \( K_{\mathrm{w}} = 3.8 \times 10^{-14} \), the \( \mathrm{H^+} \) concentration becomes \( 6.1644 \times 10^{-8} \text{ M} \).
- \( [\mathrm{H^+}] = \sqrt{K_{\mathrm{w}}} \)
Calculated concentrations of ions are crucial in determining the pH of solutions. For example, at a temperature of 40°C with \( K_{\mathrm{w}} = 3.8 \times 10^{-14} \), the \( \mathrm{H^+} \) concentration becomes \( 6.1644 \times 10^{-8} \text{ M} \).
- Knowing the \( \mathrm{H^+} \) concentration, the pH is calculated as \(-\log(\mathrm{H^+})\).