How many mL of 4.50M sodium hydroxide must be acetic acid solutions to make a buffer with pHb. How many mL of 4.50 M sodium hydroxide must be added to 250.0 mL of a 0.200 Mc. Calculate the pH of a formic acid solution that contains 1.45% formic acid by mass.
Question 4a. The pH of the acid and its conjugate base pairbuffer solution can be calculated by using(Assume a density of 1.01 g/mL for the solution.)acetic acid solution to make a buffer with pH = 5.000?the Anderson Hassel Balch equation:=5.000
The correct answer and explanation is :
To address the problems, let’s break them down one by one.
1. pH of a buffer solution (Acetic acid + Sodium hydroxide)
We can use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution containing a weak acid and its conjugate base:
[
\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
]
Where:
- pKa is the acid dissociation constant of the weak acid (acetic acid here).
- ([A^-]) is the concentration of the conjugate base (acetate ion).
- ([HA]) is the concentration of the weak acid (acetic acid).
Given Data for Acetic Acid:
- The pKa of acetic acid is 4.76.
- Desired pH is 5.00.
- You are asked to find how many mL of 4.50M sodium hydroxide is needed to achieve this pH.
For acetic acid, adding sodium hydroxide will neutralize some of the acid to form its conjugate base, acetate. Let’s calculate the ratio of the concentrations of the base and the acid in the Henderson-Hasselbalch equation:
[
5.00 = 4.76 + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
]
Rearranging the equation:
[
\log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) = 5.00 – 4.76 = 0.24
]
So,
[
\frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.24} \approx 1.737
]
This means the ratio of the concentration of the acetate ion (A⁻) to acetic acid (HA) should be about 1.737.
Volume of Sodium Hydroxide to Add:
Now, assuming you have 250.0 mL of a 0.200 M acetic acid solution, and you are adding sodium hydroxide (NaOH), which dissociates completely to form OH⁻ ions. Each OH⁻ ion will neutralize one H⁺ ion from the acetic acid to form acetate ions.
First, let’s calculate how many moles of acetic acid (HA) are present:
[
\text{moles of HA} = 0.200 \, \text{M} \times 0.250 \, \text{L} = 0.0500 \, \text{moles}
]
Let x be the number of moles of NaOH added (which is equal to the moles of acetate formed). The moles of acetic acid left will be:
[
\text{moles of HA} = 0.0500 – x
]
The moles of acetate formed will be equal to the moles of NaOH added, so:
[
\text{moles of A}^- = x
]
Now, from the Henderson-Hasselbalch equation, we know:
[
\frac{x}{0.0500 – x} = 1.737
]
Solving for x:
[
x = 1.737 \times (0.0500 – x)
]
[
x = 1.737 \times 0.0500 – 1.737x
]
[
x + 1.737x = 0.08685
]
[
2.737x = 0.08685
]
[
x = \frac{0.08685}{2.737} \approx 0.0318 \, \text{moles of NaOH}
]
Now, to find the volume of 4.50 M NaOH solution required:
[
\text{volume of NaOH} = \frac{\text{moles of NaOH}}{\text{concentration of NaOH}} = \frac{0.0318}{4.50} \approx 0.00707 \, \text{L} = 7.07 \, \text{mL}
]
Thus, you need to add approximately 7.07 mL of 4.50 M NaOH to 250.0 mL of a 0.200 M acetic acid solution to achieve a pH of 5.00.
2. pH of a Formic Acid Solution
Given that formic acid has a concentration of 1.45% by mass, we can calculate the pH of the solution. First, let’s assume we have 100 grams of solution (for simplicity).
The mass of formic acid (HCOOH) in the solution would be:
[
\text{Mass of HCOOH} = 1.45 \, \text{g}
]
Next, we need to calculate the moles of formic acid. The molar mass of formic acid is approximately 46.03 g/mol:
[
\text{moles of HCOOH} = \frac{1.45 \, \text{g}}{46.03 \, \text{g/mol}} \approx 0.0315 \, \text{mol}
]
Now, calculate the volume of the solution. We assume a density of 1.01 g/mL:
[
\text{volume of solution} = \frac{\text{mass of solution}}{\text{density}} = \frac{100 \, \text{g}}{1.01 \, \text{g/mL}} \approx 99.01 \, \text{mL} = 0.09901 \, \text{L}
]
Now, calculate the molarity of formic acid:
[
\text{Molarity of HCOOH} = \frac{0.0315 \, \text{mol}}{0.09901 \, \text{L}} \approx 0.318 \, \text{M}
]
Formic acid has a pKa of 3.75, and its dissociation in water is:
[
\text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^-
]
For a weak acid, we can use the ICE table and the Ka expression to calculate the concentration of H⁺ ions:
[
\text{Ka} = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}
]
[
\text{Ka} = 10^{-3.75} = 1.78 \times 10^{-4}
]
Using the approximation for weak acids (since the dissociation is small), we can find:
[
[\text{H}^+] \approx \sqrt{\text{Ka} \times [\text{HCOOH}]} = \sqrt{1.78 \times 10^{-4} \times 0.318} \approx 7.53 \times 10^{-3} \, \text{M}
]
Finally, calculate the pH:
[
\text{pH} = -\log(7.53 \times 10^{-3}) \approx 2.12
]
Thus, the pH of the formic acid solution is approximately 2.12.
3. The Anderson-Hasselbalch Equation
This equation is another way of calculating the pH of a buffer solution, similar to the Henderson-Hasselbalch equation, but it specifically deals with more complex equilibrium systems where both weak acid and conjugate base are present in significant amounts. The form of the equation is:
[
\text{pH} = \text{pKa} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)
]
It is particularly useful when you have a mixture of a weak acid and its conjugate base, and it can help fine-tune buffer solution preparation.