Problem 26
Question
How many grams of sodium lactate \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COONa}\right.\) or \(\left.\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) should be added to \(1.00 \mathrm{~L}\) of \(0.150 \mathrm{M}\) lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to form a buffer solution with \(\mathrm{pH} 4.00\) ? Assume that no volume change occurs when the sodium lactate is added.
Step-by-Step Solution
Verified Answer
To form a buffer solution with pH 4.00, add 16.8 grams of sodium lactate to 1.00 L of 0.150 M lactic acid. This is calculated using the Henderson-Hasselbalch equation and the given pH and molar concentration of lactic acid. The concentration of sodium lactate is also found to be 0.150 M in the buffer solution.
1Step 1: Determine the pKa of lactic acid
To find the pKa of lactic acid, we can use the given pH of the buffer solution. We know that pH = pKa when the concentration of the acid (HA) and its conjugate base (A-) are equal. In this case, the pH is given as 4.00, so the pKa of lactic acid will also be 4.00.
2Step 2: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation can be used to determine the concentration of the conjugate base (sodium lactate) required for the buffer solution:
pH = pKa + log(\(\frac{[A^-]}{[HA]}\))
Where pH is the desired pH of the buffer, pKa is the pKa value of the weak acid, [A-] is the molar concentration of the conjugate base (sodium lactate), and [HA] is the molar concentration of the weak acid (lactic acid). Plugging in the known values, we get:
4.00 = 4.00 + log(\(\frac{[A^-]}{0.150}\))
3Step 3: Solve for [A-]
Now we need to solve for the concentration of the conjugate base (sodium lactate) in the solution. From the previous step, the equation becomes:
0 = log(\(\frac{[A^-]}{0.150}\))
To remove the logarithm, we can use the inverse function, which is the exponentiation:
\(10^0\) = \(\frac{[A^-]}{0.150}\)
1 = \(\frac{[A^-]}{0.150}\)
[A-] = (1)(0.150) = 0.150 M
4Step 4: Calculate the grams of sodium lactate
Now that we have the concentration of sodium lactate in the solution, we can convert it to grams using its molar mass. The molar mass of sodium lactate is 112 g/mol:
grams of sodium lactate = (0.150 mol/L)(1 L)(112 g/mol) = 16.8 g
Therefore, 16.8 grams of sodium lactate should be added to 1.00 L of 0.150 M lactic acid to form a buffer solution with pH 4.00.
Key Concepts
Henderson-Hasselbalch EquationpH CalculationMolar ConcentrationWeak Acid and Conjugate Base Equilibrium
Henderson-Hasselbalch Equation
Understanding the Henderson-Hasselbalch equation is crucial for anyone studying chemistry, especially when it comes to preparing buffer solutions. This equation provides a straightforward relationship between the pH of a buffer solution and the molar concentrations of an acid and its conjugate base.
A buffer solution resists changes in pH when small amounts of an acid or a base are added. This is possible because it contains a weak acid and its conjugate base in equilibrium. The Henderson-Hasselbalch equation is expressed as:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
Here, pH is the measure of the acidity or basicity of the solution. pKa is the acid dissociation constant, a value that indicates the strength of the weak acid. The ratio \( \frac{[\text{A}^-]}{[\text{HA}]} \) represents the molar concentrations of the conjugate base (\( \text{A}^- \)) to the weak acid (\( \text{HA} \)). By using this equation, one can predict the pH of the buffer solution or adjust concentrations to achieve a desired pH.
To apply the equation effectively, it's important to remember a few key tips:
A buffer solution resists changes in pH when small amounts of an acid or a base are added. This is possible because it contains a weak acid and its conjugate base in equilibrium. The Henderson-Hasselbalch equation is expressed as:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
Here, pH is the measure of the acidity or basicity of the solution. pKa is the acid dissociation constant, a value that indicates the strength of the weak acid. The ratio \( \frac{[\text{A}^-]}{[\text{HA}]} \) represents the molar concentrations of the conjugate base (\( \text{A}^- \)) to the weak acid (\( \text{HA} \)). By using this equation, one can predict the pH of the buffer solution or adjust concentrations to achieve a desired pH.
To apply the equation effectively, it's important to remember a few key tips:
- Ensure that you have accurate values for the pKa of the weak acid you're using.
- Always express the molar concentrations of the acid and its conjugate base in the same units before calculating.
- Understand that the closer the ratio of \( \text{A}^- \) to \( \text{HA} \) is to 1, the more effective the buffer will be at maintaining its pH.
pH Calculation
pH calculation is a fundamental skill in chemistry, pivotal to various applications including the creation of buffer solutions, pharmaceuticals, and the food industry. The pH scale ranges from 0 to 14; 7 is considered neutral, below 7 is acidic, and above 7 is basic.
The pH of a solution can be calculated using the following formula:
\[ \text{pH} = -\log([\text{H}^+]) \]
Where \( [\text{H}^+] \) represents the molar concentration of hydrogen ions. In the context of buffer solutions, the molar concentrations of the weak acid and its conjugate base play a critical role in determining the overall pH, as seen in the Henderson-Hasselbalch equation.
When using this equation for buffer solutions, one often has to operate in reverse, adjusting the pH by manipulating the concentrations of the weak acid and conjugate base. The goal is to achieve a balance that will hold the pH steady at the desired value.
To enhance clarity in pH calculations:
The pH of a solution can be calculated using the following formula:
\[ \text{pH} = -\log([\text{H}^+]) \]
Where \( [\text{H}^+] \) represents the molar concentration of hydrogen ions. In the context of buffer solutions, the molar concentrations of the weak acid and its conjugate base play a critical role in determining the overall pH, as seen in the Henderson-Hasselbalch equation.
When using this equation for buffer solutions, one often has to operate in reverse, adjusting the pH by manipulating the concentrations of the weak acid and conjugate base. The goal is to achieve a balance that will hold the pH steady at the desired value.
To enhance clarity in pH calculations:
- Be mindful of the sign; the logarithm of a number less than one is negative, and since pH is typically a positive number, the negative sign in the equation is essential.
- Remember that pH is a logarithmic scale, meaning each whole number change represents a tenfold increase or decrease in acidity.
- Accuracy in measuring the concentration of \( [\text{H}^+] \) is crucial to obtaining correct pH values.
Molar Concentration
Molar concentration, often denoted as molarity and expressed as moles per liter (mol/L), is another key concept in understanding buffer solutions. It quantifies the amount of a solute -- in this case, the weak acid or its conjugate base -- within a given volume of solution.
The molar concentration is calculated by dividing the number of moles of the solute by the volume of the solution in liters:
\[ \text{Molar concentration} (\text{M}) = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
For buffer solution preparation, it's essential to know the molar concentrations of both the weak acid and its conjugate base. This is because the effectiveness of the buffer relies on their ratio, as represented in the Henderson-Hasselbalch equation.
Key points to consider when dealing with molar concentration:
The molar concentration is calculated by dividing the number of moles of the solute by the volume of the solution in liters:
\[ \text{Molar concentration} (\text{M}) = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
For buffer solution preparation, it's essential to know the molar concentrations of both the weak acid and its conjugate base. This is because the effectiveness of the buffer relies on their ratio, as represented in the Henderson-Hasselbalch equation.
Key points to consider when dealing with molar concentration:
- Accuracy in measurement is vital -- ensure that volumes and masses are measured precisely.
- Always use the same unit for volume when comparing concentrations to maintain consistency.
- Consider the impact of dilution or evaporation on the solution's volume, as this will affect concentration and thus the buffer's ability to stabilize pH.
Weak Acid and Conjugate Base Equilibrium
The concept of equilibrium between a weak acid and its conjugate base is at the heart of buffer systems. A weak acid partially dissociates in solution to produce hydrogen ions (H+) and its conjugate base (A-). The equilibrium can be represented as follows:
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]
In a buffer solution, the presence of both the weak acid and the conjugate base allows the solution to resist changes in pH. These components neutralize small amounts of added acids or bases. This dynamic equilibrium is central to understanding how buffers work and how to prepare them.
When considering this equilibrium, remember:
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]
In a buffer solution, the presence of both the weak acid and the conjugate base allows the solution to resist changes in pH. These components neutralize small amounts of added acids or bases. This dynamic equilibrium is central to understanding how buffers work and how to prepare them.
When considering this equilibrium, remember:
- The strength of a weak acid is defined by its dissociation constant (Ka), and pKa is its negative logarithm. Lower pKa values indicate stronger acids.
- A buffer performs optimally when the pH is close to the pKa of the weak acid and the ratio of the conjugate base to the weak acid is about 1:1.
- Le Chatelier's Principle states that the system at equilibrium will act to counteract the effects of any disturbance, which is why buffer solutions can neutralize added acids or bases without a significant change in pH.
Other exercises in this chapter
Problem 23
A buffer is prepared by adding \(20.0 \mathrm{~g}\) of acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) and \(20.0 \mathrm{~g}\) of sodium acetate \(\
View solution Problem 24
A buffer is prepared by adding \(7.00 \mathrm{~g}\) of ammonia \(\left(\mathrm{NH}_{3}\right)\) and \(20.0 \mathrm{~g}\) of ammonium chloride \(\left(\mathrm{NH
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A buffer solution contains \(0.10 \mathrm{~mol}\) of acetic acid and $0.13 \mathrm{~mol}\( of sodium acetate in \)1.00 \mathrm{~L}$. (a) What is the \(\mathrm{p
View solution Problem 29
(a) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood of pH \(7.4\) ? (b) What is the ratio of \(\mathrm{HCO}_{3}^{-}
View solution