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 your question, we need to calculate the amount of sodium hydroxide (NaOH) required to adjust the pH of an acetic acid solution to 5.000, as well as calculate the pH of a formic acid solution with a specific concentration.
Part 1: Buffer Preparation Using Sodium Hydroxide for Acetic Acid
To prepare a buffer with pH 5.000, you will use the Henderson-Hasselbalch equation:
[
\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
]
Where:
- pKa of acetic acid = 4.76
- pH desired = 5.00
- [A⁻] = concentration of acetate ions (the conjugate base of acetic acid)
- [HA] = concentration of acetic acid
The equation can be rearranged to solve for the ratio of [A⁻] to [HA]:
[
\text{pH} – \text{pKa} = \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
]
[
\log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) = 5.00 – 4.76 = 0.24
]
[
\frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.24} \approx 1.737
]
This means that for every 1 mole of acetic acid, there should be 1.737 moles of acetate (A⁻).
Now, if you start with a 0.200 M acetic acid solution and wish to add NaOH, NaOH will neutralize some of the acetic acid, converting it to acetate (A⁻). The reaction is:
[
\text{CH₃COOH} + \text{OH}⁻ \rightarrow \text{CH₃COO}⁻ + \text{H₂O}
]
To achieve the desired ratio, we need to calculate how many moles of NaOH are required. For a 250 mL solution (0.250 L) of 0.200 M acetic acid:
[
\text{Moles of CH₃COOH} = 0.200 \, \text{M} \times 0.250 \, \text{L} = 0.050 \, \text{moles}
]
From the ratio of 1.737, the moles of acetate needed are:
[
\text{Moles of CH₃COO}^- = 1.737 \times 0.050 = 0.087 \, \text{moles}
]
Since NaOH is a strong base and reacts in a 1:1 ratio with acetic acid, the moles of NaOH required are equal to the moles of acetate formed:
[
\text{Moles of NaOH} = 0.087 \, \text{moles}
]
Now, calculate the volume of 4.50 M NaOH needed to provide 0.087 moles:
[
\text{Volume of NaOH} = \frac{\text{Moles of NaOH}}{\text{Concentration of NaOH}} = \frac{0.087 \, \text{moles}}{4.50 \, \text{M}} = 0.0193 \, \text{L} = 19.3 \, \text{mL}
]
Thus, 19.3 mL of 4.50 M NaOH should be added to 250.0 mL of 0.200 M acetic acid to achieve a buffer with pH = 5.000.
Part 2: pH of a Formic Acid Solution
To calculate the pH of a formic acid solution containing 1.45% formic acid by mass, we need to use the Henderson-Hasselbalch equation again. First, let’s determine the concentration of formic acid.
The mass percentage of formic acid is given as 1.45%, so in 100 g of solution, there are 1.45 g of formic acid. Assuming the density of the solution is 1.01 g/mL, the volume of 100 g of solution is:
[
\text{Volume of solution} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \, \text{g}}{1.01 \, \text{g/mL}} \approx 99.0 \, \text{mL} = 0.0990 \, \text{L}
]
The moles of formic acid (HCOOH) in 1.45 g are:
[
\text{Moles of HCOOH} = \frac{1.45 \, \text{g}}{\text{Molar mass of HCOOH} (46.03 \, \text{g/mol})} = 0.0315 \, \text{moles}
]
Thus, the concentration of formic acid in the solution is:
[
\text{Concentration of HCOOH} = \frac{0.0315 \, \text{moles}}{0.0990 \, \text{L}} = 0.318 \, \text{M}
]
The pKa of formic acid is 3.75. Since the solution is primarily formic acid, we assume there is little dissociation, so the concentration of formate ions (HCOO⁻) is very small.
Using the Henderson-Hasselbalch equation and the fact that the concentration of formate ions is negligible, the pH of the formic acid solution will approximately be the pKa:
[
\text{pH} \approx \text{pKa of HCOOH} = 3.75
]
Thus, the pH of the formic acid solution is approximately 3.75.
Summary
- To prepare a buffer with pH 5.000 from a 0.200 M acetic acid solution, you need to add 19.3 mL of 4.50 M NaOH.
- The pH of a formic acid solution containing 1.45% formic acid by mass is approximately 3.75.