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 :
Let’s break down the questions and solve each part step by step:
1. How many mL of 4.50 M sodium hydroxide must be added to 250.0 mL of a 0.200 M acetic acid solution to make a buffer with pH = 5.00?
To calculate how much sodium hydroxide (NaOH) is required, we can use the Henderson-Hasselbalch equation:
[
\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
]
Where:
- pH = 5.00
- pKa of acetic acid = 4.76 (common value)
- ([\text{A}^-]) is the concentration of the acetate ion (CH₃COO⁻), the conjugate base
- ([\text{HA}]) is the concentration of acetic acid (CH₃COOH)
We want to find the ratio ([\text{A}^-]/[\text{HA}]) that gives us a pH of 5.00.
Rearranging the equation:
[
\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)
]
Now, solving for the ratio:
[
\frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.24} \approx 1.737
]
Next, we need to determine how much NaOH (sodium hydroxide) to add to the acetic acid solution to achieve this ratio. NaOH will convert acetic acid (HA) into acetate ion (A⁻). The moles of NaOH required are related to the moles of acetic acid that will be converted to acetate.
2. Calculation of the volume of NaOH required:
- Initially, we have 250.0 mL of a 0.200 M acetic acid solution. The moles of acetic acid are:
[
\text{moles of HA} = 0.200 \, \text{M} \times 0.250 \, \text{L} = 0.0500 \, \text{mol}
]
- To reach the ratio ([\text{A}^-]/[\text{HA}] = 1.737), we need to have a certain amount of acetate ion (A⁻), so the moles of acetate ion required are:
[
\text{moles of A}^- = 1.737 \times \text{moles of HA} = 1.737 \times 0.0500 \, \text{mol} = 0.08685 \, \text{mol}
]
- The amount of NaOH needed is equal to the moles of acetate formed, so we need 0.08685 moles of NaOH. Using the concentration of NaOH (4.50 M):
[
\text{volume of 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 is required to make the buffer with a pH of 5.00.
3. Calculate the pH of a formic acid solution that contains 1.45% formic acid by mass.
The formula for formic acid is HCOOH, and its dissociation constant (Ka) is approximately (1.77 \times 10^{-4}).
To find the pH of the formic acid solution, we will follow these steps:
- Calculate the molarity of formic acid in the solution:
The density of the solution is 1.01 g/mL, and we have 1.45% formic acid by mass. This means 1.45 g of formic acid in 100 g of solution. For 1 L of solution (1000 mL), the mass of the solution is:
[
\text{mass of solution} = 1.01 \, \text{g/mL} \times 1000 \, \text{mL} = 1010 \, \text{g}
]
The mass of formic acid in 1 L of solution is:
[
\text{mass of formic acid} = 1.45\% \times 1010 \, \text{g} = 14.645 \, \text{g}
]
Now, convert the mass of formic acid to moles:
[
\text{moles of formic acid} = \frac{14.645 \, \text{g}}{46.03 \, \text{g/mol}} = 0.318 \, \text{mol}
]
The molarity of formic acid is:
[
\text{Molarity of HCOOH} = \frac{0.318 \, \text{mol}}{1 \, \text{L}} = 0.318 \, \text{M}
]
- Set up the equilibrium expression for formic acid dissociation:
[
\text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^-
]
Let (x) be the concentration of H⁺ (and HCOO⁻) at equilibrium. The Ka expression is:
[
K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}
]
Substituting the values:
[
K_a = \frac{x^2}{0.318 – x}
]
[
1.77 \times 10^{-4} = \frac{x^2}{0.318 – x}
]
Assuming (x) is small compared to 0.318 M, we approximate:
[
1.77 \times 10^{-4} \approx \frac{x^2}{0.318}
]
Solving for (x):
[
x^2 = (1.77 \times 10^{-4}) \times 0.318 = 5.63 \times 10^{-5}
]
[
x = \sqrt{5.63 \times 10^{-5}} = 7.51 \times 10^{-3} \, \text{M}
]
- Calculate the pH:
The pH is given by:
[
\text{pH} = -\log [\text{H}^+] = -\log (7.51 \times 10^{-3}) \approx 2.12
]
Thus, the pH of the formic acid solution is approximately 2.12.
Conclusion:
- To create a buffer with pH = 5.00 using acetic acid and sodium hydroxide, 19.3 mL of 4.50 M NaOH is required.
- The pH of the formic acid solution with 1.45% formic acid by mass is approximately 2.12.