Problem 13 - Analytical Chemistry

Question

Calculate the hydrogen ion concentration and pH of a neutral solution at \(50^{\circ} \mathrm{C}\left(K_{w}=5.5 \times 10^{-14}\right.\) at \(50^{\circ} \mathrm{C}\) ).

Step-by-Step Solution

Answer

[H⁺] = 2.35 × 10⁻⁷ M, pH = 6.63 at 50°C.

Step 1: Understanding the Problem

We need to calculate the hydrogen ion concentration \([H^+]\) and the pH of a neutral solution where the ion product of water \(K_w\) is given as \(5.5 \times 10^{-14}\) at \(50^{\circ} \mathrm{C}\). A neutral solution means \([H^+] = [OH^-]\).

Step 2: Apply Ion Product of Water

For a neutral solution, the ion product of water \(K_w\) equals the product of hydrogen ion concentration \([H^+]\) and hydroxide ion concentration \([OH^-]\). Since \([H^+] = [OH^-]\) in a neutral solution, we have \(K_w = [H^+]^2\).

Step 3: Calculate Hydrogen Ion Concentration

We set up the equation \([H^+]^2 = 5.5 \times 10^{-14}\) and solve for \([H^+]\) by taking the square root: \([H^+] = \sqrt{5.5 \times 10^{-14}}\).

Step 4: Solve for [H⁺]

Calculate \([H^+] = \sqrt{5.5 \times 10^{-14}} = 2.35 \times 10^{-7}\) M. Thus, the hydrogen ion concentration \([H^+]\) is \(2.35 \times 10^{-7}\) moles per liter.

Step 5: Calculate pH

Use the pH formula: \(\mathrm{pH} = -\log_{10}([H^+])\). Plug in the value of \([H^+]\) to get \(\mathrm{pH} = -\log_{10}(2.35 \times 10^{-7})\).

Step 6: Find the pH Value

Calculate the pH: \(\mathrm{pH} = 6.63\). Therefore, the pH of the solution at \(50^{\circ} \mathrm{C}\) is 6.63.

Key Concepts

Hydrogen Ion Concentration

In chemistry, understanding hydrogen ion concentration is crucial. The concentration of hydrogen ions, denoted as \([H^+]\), plays a key role in determining the acidity or basicity of a solution. \([H^+]\) is expressed in moles per liter (M). To calculate \([H^+]\), especially in a neutral solution, you need to understand its relationship with the ion product of water (\(K_w\)). At standard conditions (25°C), \([H^+]\) is equal to \([OH^-]\) meaning \([H^+] = 1.0 \times 10^{-7}\) M. However, this can change with temperature. To find \([H^+]\) at different temperatures, you set up the equation \([H^+]^2 = K_w\). Solving this gives you the concentration of hydrogen ions, which is essential for calculating pH.

Ion Product of Water

The ion product of water, \(K_w\), is a constant that represents the product of hydrogen ions and hydroxide ions concentrations in water. It varies with temperature and is fundamental to understanding water's behavior as a solvent. At 25°C, \(K_w = 1.0 \times 10^{-14}\) M², but it increases with temperature. In the given problem, \(K_w\) at 50°C is \(5.5 \times 10^{-14}\) M². When \(K_w\) changes, the nature of neutrality changes too. For a neutral solution, \([H^+] = [OH^-]\), so \(K_w = [H^+]^2\). By knowing \(K_w\), you can solve for \([H^+]\), and thus, determine the pH of the solution.

Neutral Solution

A neutral solution is one where the number of hydrogen ions equals the number of hydroxide ions, \([H^+] = [OH^-]\). At 25°C, a neutral solution has a pH of 7, corresponding to \([H^+] = 1.0 \times 10^{-7}\) M. However, the neutrality concept remains the same even if the pH isn’t exactly 7 at other temperatures.When temperature changes, \(K_w\) varies, altering both \([H^+]\) and pH. For instance, at 50°C, a neutral solution has \([H^+]\) calculated to be \(2.35 \times 10^{-7}\) M, resulting in a pH of 6.63, which is still neutral because \([H^+] = [OH^-]\).Thus, neutrality isn't fixed to pH 7 but depends on \(K_w\) at the given temperature. This concept underlies many practical applications and pH adjustments for various biological and chemical processes.