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 your question step by step and clarify the concepts involved.
Part 1: Amount of Sodium Hydroxide Required for a Buffer Solution
You want to know how many mL of 4.50 M sodium hydroxide (NaOH) must be added to a 250.0 mL acetic acid solution to create a buffer with a pH of 5.000.
- Buffering with Acetic Acid (CH₃COOH) and its Conjugate Base (CH₃COO⁻):
The key to buffer preparation is to mix a weak acid (acetic acid) with its conjugate base (acetate, CH₃COO⁻). The pH of such a buffer can be calculated using the Henderson-Hasselbalch equation: [
\text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)
] Where:
- pKa of acetic acid = 4.76
- pH is given as 5.00.
- Determine the ratio of base (acetate) to acid (acetic acid):
First, rearrange the equation to solve for the ratio of the conjugate base (acetate) to the weak acid (acetic acid): [
\text{pH} – \text{pKa} = \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)
] [
5.00 – 4.76 = \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)
] [
0.24 = \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)
] [
10^{0.24} \approx 1.737
] Therefore, the ratio of base (acetate) to acid (acetic acid) needs to be approximately 1.737. - Stoichiometry:
When NaOH is added to acetic acid, it neutralizes part of the acetic acid to form acetate. The amount of NaOH needed is directly related to the amount of acetic acid and the desired ratio of conjugate base to acid. Suppose the initial amount of acetic acid is 0.200 M, and you have 250.0 mL of this solution. The moles of acetic acid are: [
\text{moles of acetic acid} = 0.200 \, \text{M} \times 0.2500 \, \text{L} = 0.0500 \, \text{mol}
] To achieve a buffer ratio of 1.737, you would need: [
\text{moles of acetate} = 1.737 \times \text{moles of acetic acid} = 1.737 \times 0.0500 \, \text{mol} = 0.08685 \, \text{mol}
] - Volume of NaOH:
To neutralize the necessary amount of acetic acid and create the required amount of acetate, we need to add NaOH. The NaOH will dissociate into OH⁻ ions, which will neutralize CH₃COOH to form CH₃COO⁻. The moles of NaOH required are equal to the moles of acetate needed, which is 0.08685 mol. Given that NaOH is 4.50 M, the volume of NaOH required is: [
\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}
] So, 19.3 mL of 4.50 M NaOH is required to prepare the buffer solution.
Part 2: pH of Formic Acid Solution
You’re also asking about the pH of a formic acid solution that contains 1.45% formic acid by mass. To calculate this, follow these steps:
- Convert Mass Percentage to Moles:
Assume you have 100 mL of the solution. The density of the solution is given as 1.01 g/mL, so the mass of the solution is: [
\text{mass of solution} = 100 \, \text{mL} \times 1.01 \, \text{g/mL} = 101 \, \text{g}
] The amount of formic acid in the solution is 1.45% by mass: [
\text{mass of formic acid} = 1.45\% \times 101 \, \text{g} = 1.465 \, \text{g}
] - Calculate the Moles of Formic Acid:
The molar mass of formic acid (HCOOH) is approximately 46.03 g/mol. Thus, the moles of formic acid are: [
\text{moles of formic acid} = \frac{1.465 \, \text{g}}{46.03 \, \text{g/mol}} = 0.0318 \, \text{mol}
] - Concentration of Formic Acid:
The volume of the solution is 100 mL, or 0.100 L. The molarity (M) of formic acid is: [
\text{Molarity} = \frac{0.0318 \, \text{mol}}{0.100 \, \text{L}} = 0.318 \, \text{M}
] - pH Calculation Using the Henderson-Hasselbalch Equation:
The pKa of formic acid is 3.75. To calculate the pH, we would typically use the Henderson-Hasselbalch equation, but since no base is added, we can instead use the Ka expression for formic acid to solve for the concentration of H⁺ ions: [
\text{Ka} = 1.77 \times 10^{-4}
] Using the approximation for weak acids: [
\text{pH} = \frac{1}{2} \left( \text{pKa} – \log [\text{HA}] \right)
] [
\text{pH} = \frac{1}{2} \left( 3.75 – \log(0.318) \right)
] [
\text{pH} \approx \frac{1}{2} \left( 3.75 – (-0.497) \right)
] [
\text{pH} \approx \frac{1}{2} \times 4.247 = 2.12
]
Thus, the pH of the formic acid solution is approximately 2.12.
Explanation of the Anderson-Hasselbalch Equation:
The Anderson-Hasselbalch equation is used to describe buffer systems. It’s a variation of the Henderson-Hasselbalch equation, which helps calculate the pH of a buffer solution based on the concentrations of the acid and its conjugate base. In this case, since you are using acetic acid and its conjugate base (acetate), the buffer solution can be adjusted to a specific pH by varying the ratio of acid and base. The equation helps in designing buffers to maintain a consistent pH, which is essential in many chemical and biological processes.