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 :
Question 1: Volume of 4.50M Sodium Hydroxide Needed to Prepare a Buffer Solution
We are tasked with determining how many milliliters (mL) of 4.50 M sodium hydroxide (NaOH) must be added to acetic acid (CH₃COOH) to make a buffer with a given pH of 5.00.
Relevant Equation: Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the ratio of the concentrations of its conjugate base (acetate, CH₃COO⁻) and its acid (acetic acid, CH₃COOH):
[
\text{pH} = \text{pKa} + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
]
where:
- pH is the desired pH of the buffer (5.00 in this case).
- pKa for acetic acid is 4.76.
- ([\text{A}^-]) is the concentration of the conjugate base (acetate).
- ([\text{HA}]) is the concentration of the acid (acetic acid).
Rearranging the equation to find the ratio of the conjugate base to acid:
[
\text{pH} – \text{pKa} = \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
]
[
5.00 – 4.76 = \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
]
[
0.24 = \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
]
[
\frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.24} \approx 1.737
]
This indicates that the ratio of acetate (conjugate base) to acetic acid is approximately 1.737.
Preparing the Buffer:
To make this buffer, you need to add NaOH to the acetic acid solution, which will convert some of the acetic acid (HA) into acetate (A⁻). Since the ratio of acetate to acetic acid needs to be 1.737, we can express this relationship in terms of the volumes and concentrations of the solutions.
Given:
- Volume of acetic acid solution = 250.0 mL.
- Concentration of acetic acid = 0.200 M.
- Concentration of NaOH = 4.50 M.
We first calculate the amount of acetic acid present:
[
\text{moles of acetic acid} = 0.200 \, \text{M} \times 0.250 \, \text{L} = 0.050 \, \text{mol}
]
Let ( V_{\text{NaOH}} ) be the volume of NaOH to be added. The moles of NaOH added will equal the moles of acetate formed, so:
[
\text{moles of acetate} = 1.737 \times \text{moles of acetic acid}
]
[
\text{moles of acetate} = 1.737 \times 0.050 = 0.08685 \, \text{mol}
]
Since NaOH dissociates completely, the moles of NaOH added will be equal to the moles of acetate formed. Now calculate the volume of NaOH required:
[
V_{\text{NaOH}} = \frac{\text{moles of NaOH}}{\text{concentration of NaOH}} = \frac{0.08685 \, \text{mol}}{4.50 \, \text{M}} = 0.0193 \, \text{L} = 19.3 \, \text{mL}
]
Thus, 19.3 mL of 4.50 M NaOH must be added to 250.0 mL of 0.200 M acetic acid to prepare a buffer with pH = 5.00.
Question 2: pH of Formic Acid Solution (1.45% by mass)
Now, let’s calculate the pH of a formic acid solution containing 1.45% by mass of formic acid.
Relevant Information:
- Formic acid (HCOOH) has a pKa of 3.75.
- Density of the solution is 1.01 g/mL.
- Assume you have 1 L (1000 mL) of solution.
Step 1: Calculate the mass of formic acid in 1.00 L of solution
[
\text{mass of formic acid} = \frac{1.45}{100} \times 1000 = 14.5 \, \text{g}
]
Step 2: Convert mass to moles
The molar mass of formic acid is approximately 46.03 g/mol, so:
[
\text{moles of formic acid} = \frac{14.5 \, \text{g}}{46.03 \, \text{g/mol}} \approx 0.315 \, \text{mol}
]
Step 3: Calculate the concentration of formic acid in the solution
[
[\text{HCOOH}] = \frac{0.315 \, \text{mol}}{1.00 \, \text{L}} = 0.315 \, \text{M}
]
Step 4: Use the Henderson-Hasselbalch equation
Assuming no conjugate base is present, we can calculate the pH using the following equation for a weak acid:
[
\text{pH} = \frac{1}{2} (\text{pKa} – \log[\text{HCOOH}])
]
[
\text{pH} = \frac{1}{2} (3.75 – \log[0.315])
]
[
\text{pH} = \frac{1}{2} (3.75 – (-0.502)) = \frac{1}{2} (4.252) = 2.13
]
So, the pH of the formic acid solution is approximately 2.13.
Conclusion
To make a buffer with pH 5.00 from acetic acid, 19.3 mL of 4.50 M sodium hydroxide must be added to 250.0 mL of 0.200 M acetic acid. The pH of a formic acid solution containing 1.45% formic acid by mass is approximately 2.13.