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 into two parts:
1. Volume of 4.50 M Sodium Hydroxide (NaOH) Needed to Prepare an Acetic Acid Buffer with pH = 5.000
Buffers are solutions that resist changes in pH upon the addition of small amounts of acids or bases. The pH of a buffer solution containing 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) ]
For acetic acid (CH₃COOH), the acid dissociation constant (Kₐ) is 1.8 × 10⁻⁵, giving a pKₐ:
[ \text{p}K_a = -\log(1.8 \times 10^{-5}) \approx 4.74 ]
Given the desired pH of 5.000:
[ 5.000 = 4.74 + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) ]
Solving for the ratio (\frac{[\text{A}^-]}{[\text{HA}]}):
[ \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) = 5.000 – 4.74 = 0.26 ]
[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{0.26} \approx 1.82 ]
This ratio indicates that the concentration of acetate ions ([A⁻]) should be 1.82 times that of acetic acid ([HA]) to achieve a pH of 5.000.
Assuming we start with 250.0 mL of 0.200 M acetic acid:
- Moles of acetic acid: [ 0.200\, \text{M} \times 0.250\, \text{L} = 0.050\, \text{moles} ]
Let ( x ) be the moles of NaOH added. NaOH reacts with acetic acid to form acetate ions:
[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} ]
At equilibrium:
- Moles of acetic acid remaining: [ 0.050 – x ]
- Moles of acetate formed: [ x ]
Using the ratio:
[ \frac{x}{0.050 – x} = 1.82 ]
Solving for ( x ):
[ x = 1.82 \times (0.050 – x) ]
[ x = 0.091 – 1.82x ]
[ x + 1.82x = 0.091 ]
[ 2.82x = 0.091 ]
[ x = \frac{0.091}{2.82} \approx 0.0323\, \text{moles} ]
To find the volume of 4.50 M NaOH required:
[ \text{Volume} = \frac{0.0323\, \text{moles}}{4.50\, \text{M}} \approx 0.00718\, \text{L} = 7.18\, \text{mL} ]
Therefore, approximately 7.18 mL of 4.50 M NaOH should be added to 250.0 mL of 0.200 M acetic acid to prepare a buffer with pH 5.000.
2. Calculating the pH of a Formic Acid Solution Containing 1.45% Formic Acid by Mass
To determine the pH of a formic acid (HCOOH) solution with 1.45% formic acid by mass and a density of 1.01 g/mL, follow these steps:
a. Calculate the Molarity of the Solution
Assume 100 g of the solution:
- Mass of formic acid: [ 1.45\% \times 100\, \text{g} = 1.45\, \text{g} ]
- Volume of the solution: [ \frac{100\, \text{g}}{1.01\, \text{g/mL}} \approx 99.01\, \text{mL} ] [ 99.01\, \text{mL} = 0.09901\, \text{L} ]
- Moles of formic acid (molar mass ≈ 46.03 g/mol): [ \frac{1.45\, \text{g}}{46.03\, \text{g/mol}} \approx 0.0315\, \text{moles} ]
- Molarity (M): [ \frac{0.0315\, \text{moles}}{0.09901\,