Chapter 16: Problem 40
How many grams of sodium acetate must be added to \(400.0 \mathrm{~mL}\) of \(0.500 \mathrm{M}\) acetic acid to prepare a \(\mathrm{pH} 4.35\) buffer?
Short Answer
Expert verified
Add approximately 6.40 grams of sodium acetate.
Step by step solution
01
Understand the Buffer Solution Components
We have a buffer solution made up of acetic acid (\( ext{CH}_3 ext{COOH} \)) and its conjugate base, sodium acetate (\( ext{CH}_3 ext{COONa} \)). We want to find the amount of sodium acetate needed to achieve a \( ext{pH} \) of 4.35.
02
Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is used for buffer solutions:\[\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]\ }\right).\]Here, \( [\text{A}^-] \) represents the acetate ion concentration, and \( [\text{HA}] \) is the acetic acid concentration. Find \( \text{pKa} \) of acetic acid, which is 4.76.
03
Calculate the Required Concentration Ratio
Since the desired \( \text{pH} \) is 4.35, rearrange the equation to find the ratio:\[4.35 = 4.76 + \log\left(\frac{[\text{A}^-]}{0.500}\right).\]Solve for the ratio\( \frac{[\text{A}^-]}{0.500}\)\: \[\log\left(\frac{[\text{A}^-]}{0.500}\right) = 4.35 - 4.76 = -0.41.\]Thus,\[\frac{[\text{A}^-]}{0.500} = 10^{-0.41} \approx 0.39.\]
04
Solve for Acetate Ion Concentration
We know now \( \frac{[\text{A}^-]}{0.500} \approx 0.39\). Therefore,\[[\text{A}^-] = 0.39 \times 0.500 = 0.195 \mathrm{~M}.\]
05
Calculate Moles of Sodium Acetate Needed
The **acetate ion concentration** corresponds to sodium acetate concentration. Calculate the moles of sodium acetate:\[\text{Moles of sodium acetate} = 0.195 \times 0.400 = 0.078 \text{ moles}.\]
06
Convert Moles to Grams
Convert moles of sodium acetate to grams using its molar mass (82.03 g/mol):\[\text{Grams of sodium acetate} = 0.078 \times 82.03 \approx 6.40 \text{ grams}.\]
07
Conclusion
To prepare a \( ext{pH} \) 4.35 buffer, add approximately 6.40 grams of sodium acetate to the solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Henderson-Hasselbalch equation
Buffer solutions are essential in maintaining a stable pH level in various chemical and biological processes. One of the key tools used to understand buffer solutions is the Henderson-Hasselbalch equation. This equation is particularly useful when working with acid-base chemistry because it relates the pH of a solution to the concentration of an acid and its conjugate base.
The equation is expressed as follows:\[\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]\ }\right).\]Here, \(\text{pH}\) represents the measure of acidity or basicity of a solution. The \(\text{pKa}\) is the negative base-10 logarithm of the acid dissociation constant (Ka) of the acid, providing insight into the strength of the acid.
In this equation, \([\text{A}^-]\) is the concentration of the conjugate base (such as acetate ions in our solution), and \([\text{HA}]\) is the concentration of the undissociated acid (such as acetic acid). By rearranging this equation, one can calculate the necessary proportions of acid and conjugate base to achieve a desired pH, making it a powerful predictive tool in the preparation of buffer solutions.
The equation is expressed as follows:\[\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]\ }\right).\]Here, \(\text{pH}\) represents the measure of acidity or basicity of a solution. The \(\text{pKa}\) is the negative base-10 logarithm of the acid dissociation constant (Ka) of the acid, providing insight into the strength of the acid.
In this equation, \([\text{A}^-]\) is the concentration of the conjugate base (such as acetate ions in our solution), and \([\text{HA}]\) is the concentration of the undissociated acid (such as acetic acid). By rearranging this equation, one can calculate the necessary proportions of acid and conjugate base to achieve a desired pH, making it a powerful predictive tool in the preparation of buffer solutions.
pH calculation
Understanding how to calculate \(\text{pH}\) is crucial in acid-base chemistry. It is a scale used to specify the acidity or basicity of an aqueous solution. The \(\text{pH}\) scale typically ranges from 0 to 14, where lower values indicate acidic solutions, higher values indicate basic solutions, and a \(\text{pH}\) of 7 is considered neutral.
Calculating \(\text{pH}\) can be straightforward when you have the necessary concentrations of the acidic and basic components. Using the Henderson-Hasselbalch equation, as previously discussed, you can substitute the known values such as the desired \(\text{pH}\), \(\text{pKa}\), and concentration of the acid to solve for the missing piece—often the concentration of the conjugate base in the solution.
For instance, in the given exercise, we wanted a buffer solution of \(\text{pH}\) 4.35 using acetic acid and sodium acetate. Given the \(\text{pKa}\) of acetic acid is 4.76, we rearranged the Henderson-Hasselbalch equation to find the concentration ratio of acetate ions to acetic acid, and eventually determined that we need 6.40 grams of sodium acetate to achieve the desired \(\text{pH}\).
Calculating \(\text{pH}\) can be straightforward when you have the necessary concentrations of the acidic and basic components. Using the Henderson-Hasselbalch equation, as previously discussed, you can substitute the known values such as the desired \(\text{pH}\), \(\text{pKa}\), and concentration of the acid to solve for the missing piece—often the concentration of the conjugate base in the solution.
For instance, in the given exercise, we wanted a buffer solution of \(\text{pH}\) 4.35 using acetic acid and sodium acetate. Given the \(\text{pKa}\) of acetic acid is 4.76, we rearranged the Henderson-Hasselbalch equation to find the concentration ratio of acetate ions to acetic acid, and eventually determined that we need 6.40 grams of sodium acetate to achieve the desired \(\text{pH}\).
acid-base chemistry
Acid-base chemistry is a fundamental area of study that involves the exploration of acids, bases, and their interactions with each other. It is essential in understanding a wide range of natural and industrial processes.
Acids are substances that can donate a proton (\(\text{H}^+\)) to another substance, while bases are substances that can accept a proton. When acids and bases react, they can neutralize each other, forming water and a salt. The strength of an acid or base is determined by its ability to donate or accept protons, which is quantified using \(\text{pKa}\) and \(\text{pKb}\), respectively.
In the context of buffers, we utilize the conjugate acid-base pair to resist changes in \(\text{pH}\). This buffering action is crucial in biological systems, such as maintaining the \(\text{pH}\) of blood, and in laboratory settings, where reactions require a constant \(\text{pH}\) to proceed correctly.
Understanding these interactions allows chemists to manipulate conditions to create solutions with specific properties, such as the buffer solution we've discussed, showcasing the practical application of acid-base concepts in achieving controlled chemical environments.
Acids are substances that can donate a proton (\(\text{H}^+\)) to another substance, while bases are substances that can accept a proton. When acids and bases react, they can neutralize each other, forming water and a salt. The strength of an acid or base is determined by its ability to donate or accept protons, which is quantified using \(\text{pKa}\) and \(\text{pKb}\), respectively.
In the context of buffers, we utilize the conjugate acid-base pair to resist changes in \(\text{pH}\). This buffering action is crucial in biological systems, such as maintaining the \(\text{pH}\) of blood, and in laboratory settings, where reactions require a constant \(\text{pH}\) to proceed correctly.
Understanding these interactions allows chemists to manipulate conditions to create solutions with specific properties, such as the buffer solution we've discussed, showcasing the practical application of acid-base concepts in achieving controlled chemical environments.