Question

At \(40 .^{\circ} \mathrm{C}\) the value of \(K_{\mathrm{w}}\) is \(2.92 \times 10^{-14}\) a. Calculate the \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in pure water at \(40^{\circ} \mathrm{C}\) b. What is the pH of pure water at \(40 .^{\circ} \mathrm{C} ?\) c. If the hydroxide ion concentration in a solution is \(0.10 M\) what is the \(\mathrm{pH}\) at \(40 .^{\circ} \mathrm{C} ?\)

Short answer

In pure water at \(40^{\circ} \mathrm{C}\), the concentrations of \([\mathrm{H}^{+}]\) and \([\mathrm{OH}^{-}]\) are both approximately \(5.40 \times 10^{-8}\, M\). The pH of pure water at this temperature is approximately \(7.27\). For a solution with a hydroxide ion concentration of \(0.10\, M\), the pH at \(40^{\circ} \mathrm{C}\) is approximately \(12.53\).

Step-by-step solution

Use the ion product of water to calculate ion concentrations in pure water

In pure water at any temperature, the concentrations of hydrogen and hydroxide ions are equal. The ion product of water, \(K_w\), is given by: \(K_w = [\mathrm{H}^{+}] [\mathrm{OH}^{-}]\) For pure water at \(40^{\circ} \mathrm{C}\), we are given \(K_w = 2.92 \times 10^{-14}\). Since \([\mathrm{H}^{+}] = [\mathrm{OH}^{-}]\), we can write the equation as: \(K_w = [\mathrm{H}^{+}]^2\)

Calculate the concentration of hydrogen and hydroxide ions in pure water

Solve the equation for \([\mathrm{H}^{+}]\) and then find \([\mathrm{OH}^{-}]\): \([\mathrm{H}^{+}] = \sqrt{K_w} = \sqrt{2.92 \times 10^{-14}} \approx 5.40 \times 10^{-8}\, M\) \([\mathrm{OH}^{-}] = [\mathrm{H}^{+}] \approx 5.40 \times 10^{-8}\, M\)

Calculate the pH of pure water at \(40^{\circ} \mathrm{C}\)

The pH is defined as the negative logarithm of the hydrogen ion concentration: pH \(= -\log([\mathrm{H}^{+}]) = -\log(5.40 \times 10^{-8}) \approx 7.27\)

Calculate the pH of the solution with a given hydroxide ion concentration

We are given that the hydroxide ion concentration in a solution is \(0.10\, M\). We can use the ion product of water, \(K_w\), to find the hydrogen ion concentration: \([\mathrm{H}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]} = \frac{2.92 \times 10^{-14}}{0.10} = 2.92 \times 10^{-13}\, M\) The pH of the solution can then be found as: pH \(= -\log([\mathrm{H}^{+}]) = -\log(2.92 \times 10^{-13}) \approx 12.53\) In summary, we have calculated the concentrations of hydrogen and hydroxide ions in pure water as \(5.40 \times 10^{-8}\, M\), the pH of pure water as \(7.27\), and the pH of the solution with a hydroxide ion concentration of \(0.10\, M\) as \(12.53\).

Key concepts

pH Calculation

Understanding the pH scale is crucial for many aspects of science, especially chemistry and biology. pH is a measure of how acidic or basic a solution is. The pH scale ranges from 0 to 14, with 7 being neutral. pH values less than 7 indicate acidic solutions, and values greater than 7 indicate basic solutions.

The pH of a solution is calculated using the formula:
\[\begin{equation}\text{pH} = -\log([\mathrm{H}^{+}])\end{equation}\]
where \([\mathrm{H}^{+}]\) represents the concentration of hydrogen ions in moles per liter (M). To determine pH, you take the negative logarithm of the hydrogen ion concentration. For instance, if the hydrogen ion concentration in pure water at \(40^\circ\mathrm{C}\) is around \(5.40 \times 10^{-8}\ M\), the pH is calculated as follows:
\[\begin{equation}\text{pH} = -\log(5.40 \times 10^{-8}) \approx 7.27\end{equation}\]
With this example, one can notice that even in pure water, the pH can slightly vary from the neutral point, depending on temperature, which affects the ion product of water (\(K_{\mathrm{w}}\)).

Hydrogen Ion Concentration

When it comes to the acid-base chemistry, the concentration of hydrogen ions, denoted as \([\mathrm{H}^{+}]\), is a defining factor for the nature of the solution. In pure water, or any neutral solution, the concentration of hydrogen ions is equal to that of hydroxide ions (\([\mathrm{OH}^{-}]\)). The relationship between these concentrations and the ion product of water (\(K_{\mathrm{w}}\)) is expressed by:
\[\begin{equation}K_{\mathrm{w}} = [\mathrm{H}^{+}] [\mathrm{OH}^{-}]\end{equation}\]

In an environment like pure water at \(40^\circ\mathrm{C}\), using the given \(K_{\mathrm{w}} = 2.92 \times 10^{-14}\), we obtain the concentration of hydrogen ions by taking the square root of the ion product:
\[\begin{equation}[\mathrm{H}^{+}] = \sqrt{K_{\mathrm{w}}} = \sqrt{2.92 \times 10^{-14}} \approx 5.40 \times 10^{-8}\, M\end{equation}\]
This tells us that even in neutral conditions, hydrogen ions are present and have quantifiable concentrations.

Hydroxide Ion Concentration

Similarly to hydrogen ions, the concentration of hydroxide ions (denoted as \([\mathrm{OH}^{-}]\)) is fundamental in describing the basicity of a solution. In pure water at equilibrium, \([\mathrm{H}^{+}]\) and \([\mathrm{OH}^{-}]\) are equal due to the ion product of water being consistent:
\[\begin{equation}[\mathrm{OH}^{-}] = [\mathrm{H}^{+}]\end{equation}\]

When given the ion product \(K_{\mathrm{w}}\) and the concentration of hydroxide ions, the concentration of hydrogen ions can be calculated and vice versa. This interdependence allows for the determination of one by knowing the other. For example, in a solution with a \([\mathrm{OH}^{-}]\) of \(0.10\ M\), the corresponding \([\mathrm{H}^{+}]\) is given by:
\[\begin{equation}[\mathrm{H}^{+}] = \frac{K_{\mathrm{w}}}{[\mathrm{OH}^{-}]} = \frac{2.92 \times 10^{-14}}{0.10} = 2.92 \times 10^{-13}\, M\end{equation}\]
This illustrates how a high concentration of hydroxide ions can lead to a low concentration of hydrogen ions, pushing the pH to reflect a basic solution.

Temperature Dependence of Kw

The value of the ion product of water (\(K_{\mathrm{w}}\)) is not a constant but varies with temperature. It's important to understand this temperature dependency, especially when dealing with precise chemical reactions and calculations.

At standard conditions, which are \(25^\circ\mathrm{C}\), \(K_{\mathrm{w}}\) is \(1 \times 10^{-14}\). However, as shown in the exercise, at \(40^\circ\mathrm{C}\), \(K_{\mathrm{w}}\) increases to \(2.92 \times 10^{-14}\). This increase affects the concentrations of hydrogen and hydroxide ions, which in turn will affect the pH of solutions at different temperatures.

It is crucial to use the correct value of \(K_{\mathrm{w}}\) when calculating pH at temperatures other than standard conditions. Overlooking the temperature impact could result in inaccurate calculations and a misinterpretation of a solution's properties.