Problem 29
Question
The \(\mathrm{pOH}\) of a solution is 9.40 at \(25^{\circ} \mathrm{C}\). Calculate the hydronium ion concentration of the solution.
Step-by-Step Solution
Verified Answer
The hydronium ion concentration is approximately \(2.51 \times 10^{-5}\) M.
1Step 1: Understand the relationship between pOH and pH
The relationship between pH and pOH is given by the equation: \( \mathrm{pH} + \mathrm{pOH} = 14 \). At \(25^{\circ} \mathrm{C} \), this equation allows us to find the \( \mathrm{pH} \) if we know the \( \mathrm{pOH} \).
2Step 2: Calculate the pH of the solution
Given \( \mathrm{pOH} = 9.40 \), use the relationship: \( \mathrm{pH} = 14 - \mathrm{pOH} = 14 - 9.40 = 4.60 \). Now we know the pH of the solution is 4.60.
3Step 3: Understand the relationship between pH and hydronium ion concentration
The pH of a solution is related to the hydronium ion concentration \( [\mathrm{H}_3\mathrm{O}^+] \) by the formula: \( \mathrm{pH} = -\log_{10}([\mathrm{H}_3\mathrm{O}^+]) \). We can use this to find the hydronium ion concentration after calculating the pH.
4Step 4: Calculate hydronium ion concentration
Rearrange the formula to solve for \([\mathrm{H}_3\mathrm{O}^+]\): \([\mathrm{H}_3\mathrm{O}^+] = 10^{-\mathrm{pH}} \). Using \( \mathrm{pH} = 4.60 \), we find \([\mathrm{H}_3\mathrm{O}^+] = 10^{-4.60} \approx 2.51 \times 10^{-5} \text{ M} \).
Key Concepts
pOHpHRelationship between pH and pOHAcid-base equilibria
pOH
The \(\mathrm{pOH}\) of a solution provides insight into its basicity. It measures the concentration of hydroxide ions (OH⁻) in a solution. The formula to calculate \(\mathrm{pOH}\) is \(\mathrm{pOH} = -\log_{10}([\mathrm{OH}^-])\). This logarithmic scale is similar to \(\mathrm{pH}\) and is used to understand the acidic or basic nature of a solution.
When dealing with \(\mathrm{pOH}\), you should know:
For the solution with a \(\mathrm{pOH}\) of 9.40, you can use these basics to find out more about its acidic counterparts, specifically the \(\mathrm{pH}\) and the hydronium ion concentration.
When dealing with \(\mathrm{pOH}\), you should know:
- The lower the \(\mathrm{pOH}\), the more basic the solution is.
- To switch between \(\mathrm{pH}\) and \(\mathrm{pOH}\), you can use the formula: \(\mathrm{pH} + \mathrm{pOH} = 14\).
- This relationship is valid in water at \(25^{\circ} \mathrm{C}\).
For the solution with a \(\mathrm{pOH}\) of 9.40, you can use these basics to find out more about its acidic counterparts, specifically the \(\mathrm{pH}\) and the hydronium ion concentration.
pH
The \(\mathrm{pH}\) of a solution is a measure of the concentration of hydrogen ions (H⁺) or hydronium ions \(\left(\mathrm{H}_3\mathrm{O}^+\right)\). It indicates how acidic or basic a solution is, using a scale from 0 to 14. The formula used is \(\mathrm{pH} = -\log_{10}([\mathrm{H}_3\mathrm{O}^+])\).
Here are some key points about \(\mathrm{pH}\):
In our example, once we calculated the \(\mathrm{pH}\) from a \(\mathrm{pOH}\) of 9.40 (which turned out to be 4.60), we could then determine that the solution is acidic. From the \(\mathrm{pH}\), calculations can be done to determine the hydronium ion concentration.
Here are some key points about \(\mathrm{pH}\):
- If the \(\mathrm{pH}\) is less than 7, the solution is acidic.
- If the \(\mathrm{pH}\) is greater than 7, the solution is basic.
- A \(\mathrm{pH}\) of 7 is considered neutral, as is the case with pure water.
In our example, once we calculated the \(\mathrm{pH}\) from a \(\mathrm{pOH}\) of 9.40 (which turned out to be 4.60), we could then determine that the solution is acidic. From the \(\mathrm{pH}\), calculations can be done to determine the hydronium ion concentration.
Relationship between pH and pOH
The relationship between \(\mathrm{pH}\) and \(\mathrm{pOH}\) is a fundamental aspect of acid-base chemistry. The equation \(\mathrm{pH} + \mathrm{pOH} = 14\) expresses this balance in any aqueous solution at \(25^{\circ} \mathrm{C}\).
Understanding this relationship helps in:
This relationship ensures that if you know one value, you can effortlessly deduce the other. For instance, with a \(\mathrm{pOH}\) of 9.40, the \(\mathrm{pH}\) calculates as 4.60. This knowledge is key in tackling a variety of chemical problems, particularly when determining the characteristics of an unknown solution.
Understanding this relationship helps in:
- Interconverting between \(\mathrm{pH}\) and \(\mathrm{pOH}\).
- Assessing the acidity or basicity of a solution quickly by knowing either value.
This relationship ensures that if you know one value, you can effortlessly deduce the other. For instance, with a \(\mathrm{pOH}\) of 9.40, the \(\mathrm{pH}\) calculates as 4.60. This knowledge is key in tackling a variety of chemical problems, particularly when determining the characteristics of an unknown solution.
Acid-base equilibria
Acid-base equilibria involve reactions that establish the balance between acids and bases in a solution. The essence of these equilibria lies in understanding the exchange of protons \(\left(\mathrm{H}^+\right)\) between acids and bases. \(\mathrm{pH}\) and \(\mathrm{pOH}\) values are central to analyzing such equilibria, detailing how acidic or basic a given solution is.
Consider the following aspects when exploring acid-base equilibria:
Being well-versed in acid-base equilibria allows one to predict the effects of adding acids or bases to solutions, further informing the pH or pOH changes. These reactions are at the heart of many natural and industrial processes, making comprehension vital for students and professionals alike.
Consider the following aspects when exploring acid-base equilibria:
- Equilibrium constant expressions denote the extent of ionization of acids or bases.
- A strong acid fully ionizes in water, while a weak acid only partially does so.
- These equilibria help predict how changing concentrations or conditions affects the solution.
Being well-versed in acid-base equilibria allows one to predict the effects of adding acids or bases to solutions, further informing the pH or pOH changes. These reactions are at the heart of many natural and industrial processes, making comprehension vital for students and professionals alike.
Other exercises in this chapter
Problem 27
Calculate the hydronium ion concentration in \(\mathrm{mol} / \mathrm{L}\) for solutions with the following \(\mathrm{pH}\) values: (a) \(2.42,\) (b) \(11.21,\)
View solution Problem 28
Calculate the hydronium ion concentration in \(\mathrm{mol} / \mathrm{L}\) for each of the following solutions: (a) a solution whose \(\mathrm{pH}\) is \(5.20,(
View solution Problem 30
Calculate the number of moles of \(\mathrm{KOH}\) in \(5.50 \mathrm{~mL}\) of a \(0.360 \mathrm{M} \mathrm{KOH}\) solution. What is the \(\mathrm{pOH}\) of the
View solution Problem 31
How much \(\mathrm{NaOH}\) (in grams) is needed to prepare \(546 \mathrm{~mL}\) of solution with a \(\mathrm{pH}\) of 10.00 at \(25^{\circ} \mathrm{C} ?\)
View solution