Chapter 15: Problem 10
At \(50^{\circ} \mathrm{C}\) the value of \(K_{\mathrm{w}}\) is \(5.5 \times 10^{-14}\). What are the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{OH}\) in a neutral solution at \(50^{\circ} \mathrm{C}\) ?
Short Answer
Expert verified
Both \([H_3O^+]\) and \([OH^-]\) are \(7.4 \times 10^{-7}\) M at 50°C.
Step by step solution
01
Understanding Kw
At a given temperature, the ion-product constant for water, denoted as \(K_w\), is the product of the molar concentrations of hydronium ions \([H_3O^+]\) and hydroxide ions \([OH^-]\) in a solution. For any neutral solution, \([H_3O^+] = [OH^-]\). At 50°C, we are given \(K_w = 5.5 \times 10^{-14}\).
02
Setting up the equation
Since the solution is neutral at 50°C, we have the equation \([H_3O^+] = [OH^-]\). Based on the definition of \(K_w\), we also know \(K_w = [H_3O^+][OH^-]\). Therefore, we can set up the equation:\[ (x)(x) = 5.5 \times 10^{-14} \] where \(x\) represents the concentration of \(H_3O^+\) and \(OH^-\).
03
Solving for the concentration
To find \(x\), solve the equation \(x^2 = 5.5 \times 10^{-14}\). Take the square root of both sides to find \(x\): \[ x = \sqrt{5.5 \times 10^{-14}} \]
04
Calculation
Calculate the square root of \(5.5 \times 10^{-14}\) to find the value of \(x\):\[ x = 7.4 \times 10^{-7} \] This is the concentration of both \([H_3O^+]\) and \([OH^-]\) in the neutral solution at 50°C.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ion-product Constant of Water
The ion-product constant of water, more commonly known as the water dissociation constant, is a key factor in understanding acid-base equilibrium. It is denoted by the symbol \(K_w\). This constant is defined for pure water and describes the equilibrium between water molecules and the ions they produce: hydronium ions \([H_3O^+]\) and hydroxide ions \([OH^-]\).
\[K_w = [H_3O^+][OH^-]\]
This equation signifies that the product of the concentrations of \([H_3O^+]\) and \([OH^-]\) is always equal to \(K_w\) at a given temperature. For example, at 25°C, \(K_w\) equals \(1.0 \times 10^{-14}\), but this constant changes with temperature. At 50°C, as given in the exercise, \(K_w\) is \(5.5 \times 10^{-14}\).
\[K_w = [H_3O^+][OH^-]\]
This equation signifies that the product of the concentrations of \([H_3O^+]\) and \([OH^-]\) is always equal to \(K_w\) at a given temperature. For example, at 25°C, \(K_w\) equals \(1.0 \times 10^{-14}\), but this constant changes with temperature. At 50°C, as given in the exercise, \(K_w\) is \(5.5 \times 10^{-14}\).
- The value of \(K_w\) increases with temperature, indicating that water ionizes more at higher temperatures.
- Knowing \(K_w\) helps us to determine if a solution is acidic or basic based on the concentrations of \([H_3O^+]\) and \([OH^-]\).
Hydronium Ion Concentration
The hydronium ion concentration \([H_3O^+]\) is crucial in the study of acids and bases. Hydronium ions are formed when hydrogen ions \((H^+)\) associate with water molecules, representing the tendency of an acid to donate protons.
In a neutral solution, \([H_3O^+]\) is equal to \([OH^-]\) because they are produced in equal amounts when water dissociates.
From the exercise, we know that the neutral condition at 50°C is represented by the equation:
\[[H_3O^+] = [OH^-] = x\]
And since \(K_w = 5.5 \times 10^{-14}\) at 50°C, the equation for the hydronium ion concentration becomes:
\[x^2 = 5.5 \times 10^{-14}\]
Solving for \(x\), we find:
\[x = \sqrt{5.5 \times 10^{-14}} \approx 7.4 \times 10^{-7}\]
This value represents the hydronium ion concentration in a neutral solution at the given temperature.
In a neutral solution, \([H_3O^+]\) is equal to \([OH^-]\) because they are produced in equal amounts when water dissociates.
From the exercise, we know that the neutral condition at 50°C is represented by the equation:
\[[H_3O^+] = [OH^-] = x\]
And since \(K_w = 5.5 \times 10^{-14}\) at 50°C, the equation for the hydronium ion concentration becomes:
\[x^2 = 5.5 \times 10^{-14}\]
Solving for \(x\), we find:
\[x = \sqrt{5.5 \times 10^{-14}} \approx 7.4 \times 10^{-7}\]
This value represents the hydronium ion concentration in a neutral solution at the given temperature.
- The concentration \([H_3O^+]\) indicates the acidity of a solution; the lower the value, the less acidic the solution.
- As temperature changes, \([H_3O^+]\) varies to maintain the product \([H_3O^+][OH^-]\) equal to \(K_w\).
Neutral Solution Chemistry
Understanding neutral solution chemistry is fundamental when dealing with acid and base reactions. A neutral solution is typically one where the concentrations of hydronium ions \([H_3O^+]\) and hydroxide ions \([OH^-]\) are equal.
This state occurs naturally in pure water where no other substances are influencing the equilibrium.
For any neutral solution, the formula is simple:
\[[H_3O^+] = [OH^-]\]
At different temperatures, however, the concentration of these ions and the value of \(K_w\) changes. At 50°C, \(K_w\) was given as \(5.5 \times 10^{-14}\). To find the concentration of ions in a neutral solution at this temperature, we calculated:
\[x = 7.4 \times 10^{-7}\]
Thus, at 50°C, a neutral solution has both hydronium and hydroxide ion concentrations of approximately \(7.4 \times 10^{-7}\) M.
This state occurs naturally in pure water where no other substances are influencing the equilibrium.
For any neutral solution, the formula is simple:
\[[H_3O^+] = [OH^-]\]
At different temperatures, however, the concentration of these ions and the value of \(K_w\) changes. At 50°C, \(K_w\) was given as \(5.5 \times 10^{-14}\). To find the concentration of ions in a neutral solution at this temperature, we calculated:
\[x = 7.4 \times 10^{-7}\]
Thus, at 50°C, a neutral solution has both hydronium and hydroxide ion concentrations of approximately \(7.4 \times 10^{-7}\) M.
- A neutral pH of 7 only exists at about 25°C; at 50°C, the pH of neutral water is slightly lower due to increased ion concentration.
- The "neutral" point shifts with temperature, indicating the dynamic nature of water chemistry with environmental changes.