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 queries, let’s break them down systematically:
1. Preparing a Buffer Solution with a Specific pH:
The pH of a buffer solution composed of a weak acid (HA) and its conjugate base (A⁻) can be calculated using the Henderson-Hasselbalch equation:
[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) ]
Where:
- pKₐ is the negative logarithm of the acid dissociation constant (Kₐ) of the weak acid.
- [( \text{A}^- )] is the concentration of the conjugate base.
- [( \text{HA} )] is the concentration of the weak acid.
To prepare a buffer solution with a desired pH, you can adjust the ratio of the concentrations of the acid and its conjugate base. This is typically achieved by mixing appropriate volumes of solutions containing the acid and its conjugate base (often as a salt).
2. Titration of Acetic Acid with Sodium Hydroxide:
When titrating acetic acid (a weak acid) with sodium hydroxide (a strong base), the equivalence point is reached when the amount of base added equals the amount of acid present, resulting in the formation of water and the acetate ion (the conjugate base of acetic acid).
The pH at the equivalence point will be greater than 7 because the acetate ion is a weak base and will hydrolyze to form hydroxide ions, making the solution slightly basic. The exact pH can be calculated using the formula:
[ \text{pH} = 7 + \frac{1}{2} \times \left( \text{p}K_b + \log \left( \frac{[\text{Acetate}^-]}{[\text{Acetic Acid}]} \right) \right) ]
Where:
- pK_b is the base dissociation constant of the acetate ion, related to the pKₐ of acetic acid by:
[ \text{p}K_b = 14 – \text{p}K_a ]
3. Calculating the pH of a Formic Acid Solution:
Formic acid (HCOOH) is a weak acid that partially dissociates in water. To calculate the pH of a formic acid solution with a given concentration, follow these steps:
- Determine the concentration of formic acid ([HCOOH]): If the solution is 1.45% formic acid by mass and has a density of 1.01 g/mL, calculate the molarity using the formula: [ \text{Molarity} = \frac{\text{mass of solute (g)}}{\text{molar mass of HCOOH (g/mol)} \times \text{volume of solution (L)}} ] Given that the molar mass of formic acid is approximately 46.03 g/mol, and assuming a 1.45% by mass solution, the calculation would be: [ \text{Molarity} = \frac{1.45 \, \text{g}}{46.03 \, \text{g/mol} \times 0.0999 \, \text{L}} \approx 0.314 \, \text{M} ]
- Set up the dissociation equilibrium: [ \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- ] The acid dissociation constant (Kₐ) for formic acid is approximately 1.8 × 10⁻⁴.
- Apply the ICE (Initial, Change, Equilibrium) method: Let x be the concentration of HCOOH that dissociates:
- Initial concentrations:
- [HCOOH] = 0.314 M
- [H⁺] = [HCOO⁻] = 0
- Change in concentrations:
- [HCOOH] decreases by x: 0.314 – x
- [H⁺] and [HCOO⁻] both increase by x: x
- Equilibrium concentrations:
- [HCOOH] = 0.314 – x
- [H⁺] = x
- [HCOO⁻] = x
- Set up the equilibrium expression: [ K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} = \frac{x \times x}{0.314 – x} ] Since Kₐ is small, assume that x is much smaller than 0.314 M, so 0.314 – x ≈ 0.314. This simplifies the equation to: [ 1.8 \times 10^{-4} = \frac{x^2}{0.314} ]
- Solve for x (the [H⁺] concentration): [ x^2 = (1.8 \times 10^{-4}) \times 0.314 \approx 5.652 \times 10^{-5} ] [ x = \sqrt{5.652 \times 10^{-5}} \approx 7.52 \times 10^{-3} \, \text{M} ]
- Calculate the pH: [ \text{pH} = -\log [\text{H}^+] = -\log (7.52 \times 10^{-