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 questions systematically:
1. Volume of 4.50 M Sodium Hydroxide Needed to Prepare an Acetic Acid Buffer with pH = 5.000
The pH of a buffer solution can be estimated using the Henderson-Hasselbalch equation:
[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) ]
For acetic acid (CH₃COOH), the acid dissociation constant (( K_a )) is ( 1.8 \times 10^{-5} ), yielding a ( \text{p}K_a ) of 4.74.
Given:
- Desired pH = 5.000
- ( \text{p}K_a ) = 4.74
- Initial concentration of acetic acid, [HA] = 0.200 M
- Concentration of sodium hydroxide (NaOH) = 4.50 M
First, calculate the required ratio of the conjugate base ([A⁻]) to the acid ([HA]):
[ 5.000 = 4.74 + \log \left( \frac{[\text{A}^-]}{0.200} \right) ]
[ 0.26 = \log \left( \frac{[\text{A}^-]}{0.200} \right) ]
[ 10^{0.26} \approx 1.82 ]
[ \frac{[\text{A}^-]}{0.200} = 1.82 ]
[ [\text{A}^-] = 1.82 \times 0.200 = 0.364 \, \text{M} ]
The amount of acetate ion ([A⁻]) produced equals the moles of NaOH added. Let ( V ) be the volume of 4.50 M NaOH required:
[ \frac{0.364 \, \text{mol/L} \times 0.250 \, \text{L}}{4.50 \, \text{mol/L}} = V ]
[ V = \frac{0.091}{4.50} = 0.0202 \, \text{L} ]
[ V = 20.2 \, \text{mL} ]
Therefore, approximately 20.2 mL of 4.50 M NaOH is required to achieve a buffer solution with a pH of 5.000.
2. Volume of 4.50 M Sodium Hydroxide Needed to Add to 250.0 mL of 0.200 M Acetic Acid
Assuming the goal is to create a buffer solution with a specific pH, the required volume of NaOH can be calculated as shown above. If the desired pH is 5.000, then 20.2 mL of 4.50 M NaOH should be added to 250.0 mL of 0.200 M acetic acid.
3. Calculating the pH of a Formic Acid Solution Containing 1.45% Formic Acid by Mass
Given:
- Mass percentage of formic acid (HCOOH) = 1.45%
- Density of the solution = 1.01 g/mL
- ( K_a ) of formic acid = ( 1.8 \times 10^{-4} )
Assuming 100 g of the solution:
- Mass of formic acid = 1.45 g
- Mass of water = 98.55 g
Volume of the solution:
[ \text{Volume} = \frac{100 \, \text{g}}{1.01 \, \text{g/mL}} \approx 99.01 \, \text{mL} ]
Moles of formic acid:
[ \text{Moles} = \frac{1.45 \, \text{g}}{46.03 \, \text{g/mol}} \approx 0.0315 \, \text{mol} ]
Molarity of formic acid:
[ \text{Molarity} = \frac{0.0315 \, \text{mol}}{0.09901 \, \text{L}} \approx 0.318 \, \text{M} ]
For a weak acid dissociation:
[ \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- ]
[ K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} ]
Assuming ( [\text{H}^+] = [\text{HCOO}^-] = x ) and ( [\text{HCOOH}] \approx 0.318 – x ):
[ 1.8 \times 10^{-4} = \frac{x^2}{0.318 – x} ]
Assuming ( x ) is small compared to 0.318:
[ x^2 \approx 1.8 \times 10^{-4} \times 0.318 ]
[ x^2 \approx 5.724 \times 10^{-5} ]
[ x \approx \sqrt{5.724 \times 10^{-5}} ]
[ x \approx 0.00756 \, \text{M} ]
[ \text{pH} = -\log(0.