Chapter 16: Problem 13
What is the \(\mathrm{pH}\) of a neutral solution at \(37^{\circ} \mathrm{C}\), where \(K_{w}\) equals \(2.5 \times 10^{-14} ?\)
Short Answer
Expert verified
The pH of the neutral solution at 37掳C is 6.80.
Step by step solution
01
Understanding the Relationship
At any given temperature, the ion product of water, denoted as \(K_w\), defines the concentrations of hydrogen ions ([H+]) and hydroxide ions ([OH-]) in a neutral solution. At 37掳C, \(K_w = [H^+][OH^-] = 2.5 \times 10^{-14}\). In a neutral solution, [H+] equals [OH-], so both concentrations will equal the square root of \(K_w\).
02
Calculating Hydrogen Ion Concentration
Since \([H^+] = [OH^-]\) in a neutral solution, the hydrogen ion concentration can be calculated with \([H^+] = \sqrt{K_w}\). Thus, \([H^+] = \sqrt{2.5 \times 10^{-14}}\).
03
Solving for [H+]
Calculate the square root of \(2.5 \times 10^{-14}\):\([H^+] = \sqrt{2.5 \times 10^{-14}} = 1.58 \times 10^{-7} \, \mathrm{mol/L}\).
04
Determining pH from Hydrogen Ion Concentration
The pH is calculated using the formula: \(\mathrm{pH} = -\log{[H^+]}\). Substitute the calculated hydrogen ion concentration:\(\mathrm{pH} = -\log{(1.58 \times 10^{-7})}\).
05
Solving for pH
Compute the logarithm:\(\mathrm{pH} = -(-6.802) = 6.80\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Neutral Solution
A neutral solution is a concept that represents the balance between hydrogen ions
([H+]) and hydroxide ions ([OH-]) in a given solution. This balance is crucial to maintain the solution's neutrality. In chemistry, when a solution is neutral, the
concentration of hydrogen ions is equal to the concentration of hydroxide ions.
At a standard temperature of 25掳C, pure water is neutral with concentrations like [H+] = [OH-] = 1 脳 10鈦烩伔 M. However, the temperature affects the ion product of water ( 碍鈧), altering neutrality dynamics. Thus, at different temperatures, like 37掳C in the exercise, [H+] and [OH-] concentrations change while remaining equal in a neutral solution.
At a standard temperature of 25掳C, pure water is neutral with concentrations like [H+] = [OH-] = 1 脳 10鈦烩伔 M. However, the temperature affects the ion product of water ( 碍鈧), altering neutrality dynamics. Thus, at different temperatures, like 37掳C in the exercise, [H+] and [OH-] concentrations change while remaining equal in a neutral solution.
Ion Product of Water
The ion product of water, symbolized as 碍鈧, is a constant that signifies the product of the molar concentrations of hydrogen ions and hydroxide ions in water.
In pure water, this product remains constant at any given temperature. For example, at 25掳C, it is typically 2.5 脳 10鈦宦光伌 mol虏/L虏. The formula is expressed as \[ K_w = [H^+][OH^-] \].
The value of 碍鈧 changes with temperature, explaining why the referenced exercise uses 碍鈧 = 2.5 脳 10鈦宦光伌 at 37掳C. This concept is vital for understanding how temperature impacts the neutral state of a solution, altering hydrogen and hydroxide ion concentrations yet keeping their product constant at 碍鈧.
In pure water, this product remains constant at any given temperature. For example, at 25掳C, it is typically 2.5 脳 10鈦宦光伌 mol虏/L虏. The formula is expressed as \[ K_w = [H^+][OH^-] \].
The value of 碍鈧 changes with temperature, explaining why the referenced exercise uses 碍鈧 = 2.5 脳 10鈦宦光伌 at 37掳C. This concept is vital for understanding how temperature impacts the neutral state of a solution, altering hydrogen and hydroxide ion concentrations yet keeping their product constant at 碍鈧.
Hydrogen Ion Concentration
Hydrogen ion concentration, represented by [H+], is a key factor in determining the acidity or neutrality of a solution. In neutral solutions, this concentration equals that of hydroxide ions at any temperature, specifically at 37掳C in the exercise.
To find [H+] in a neutral solution, use the ion product equation: \[ [H^+] = \sqrt{K_w} \].
In the example provided, 碍鈧=2.5 脳 10鈦宦光伌, we calculate [H+] with
To find [H+] in a neutral solution, use the ion product equation: \[ [H^+] = \sqrt{K_w} \].
In the example provided, 碍鈧=2.5 脳 10鈦宦光伌, we calculate [H+] with
- \[ [H^+] = \sqrt{2.5 脳 10鈦宦光伌} \]
- This equals \[1.58 脳 10鈦烩伔\] mol/L.
Logarithmic Calculation
Logarithmic calculations are invaluable when determining the pH of a solution, stemming from its definition as the negative logarithm of hydrogen ion concentration: \[ ext{pH} = -\log{[H^+]} \].
This scale simplifies expressing large variations in hydrogen ion concentrations. In the exercise scenario, having calculated [H+] as 1.58 脳 10鈦烩伔 mol/L, the next step involves computing its negative logarithm:
This scale simplifies expressing large variations in hydrogen ion concentrations. In the exercise scenario, having calculated [H+] as 1.58 脳 10鈦烩伔 mol/L, the next step involves computing its negative logarithm:
- \[ ext{pH} = -\log{(1.58 脳 10^{-7})}= 6.80 \]