I Have Calculated The Answers To Steps A, B, And C

I Have Calculated The Answers To Steps A B C I Need Help With Ste

I have calculated the answers to Steps A, B & C. I need help with Step D only. Please answer STEP D 1. When 50.0 mL of 0.50 M formic acid is titrated with 0.50 M sodium hydroxide, calculate the pH: a) before any of the titrant (NaOH solution) is added, (Answer to this step is pH = 2.04) b) after 15.0 mL of the titrant has been added, (Answer to this step is pH = 3.40) c) when half of the formic acid has been neutralized, and (Answer to this step is pH = 3.77) d) at the equivalence point (be sure to account for the change in volume at each step) and select a good indicator for this titration.

Paper For Above instruction

The titration of formic acid (a weak acid) with sodium hydroxide (a strong base) is a classic example illustrating acid-base equilibria and the use of indicators in titrations. Understanding how the pH changes throughout the titration enables the determination of the equivalence point and selection of an appropriate indicator. This discussion will focus on calculating the pH at the equivalence point, considering volume changes during titration, and recommending a suitable indicator.

Calculating the pH at the Equivalence Point in Titration of Formic Acid with NaOH

Given the initial concentrations and volumes, the titration involves the following parameters:

• Volume of formic acid solution: 50.0 mL (0.050 L)
• Concentration of formic acid: 0.50 M
• Concentration of NaOH: 0.50 M
• Volume of NaOH needed to reach equivalence point: 50.0 mL (since the molarities are equal and volumes are equal, the titrant volume at equivalence is also 50.0 mL)

Step-by-Step Calculation of pH at the Equivalence Point

1. Total volume at equivalence point: sum of initial acid volume and added titrant volume.

Initial acid volume: 50.0 mL. Titrant volume to reach equivalence: 50.0 mL. Therefore, the total volume:

Total volume = 50.0 mL + 50.0 mL = 100.0 mL (0.100 L)

2. Moles of formic acid initially present:

n_acid = concentration × volume = 0.50 mol/L × 0.050 L = 0.025 mol

3. Moles of NaOH added at equivalence:

n_base = 0.50 mol/L × 0.050 L = 0.025 mol

4. During titration, the formic acid is completely neutralized to form its conjugate base, formate (HCOO^-), which is a weak base. The amount of conjugate base formed:

Because all acid is neutralized, moles of formate = initial moles of formic acid = 0.025 mol

5. Concentration of formate at the equivalence point:

[HCOO^-] = 0.025 mol / 0.100 L = 0.25 M

6. Determine the pH based on the hydrolysis of the formate ion. The hydrolysis equilibrium is:

HCOO^- + H₂O ⇌ H⁺ + HCO₃^-

Using the K_b of the conjugate base:

K_b = K_w / K_a

Given that the acid dissociation constant (K_a) for formic acid is approximately 1.78 × 10^-4 (at 25°C), and that K_w = 1.0 × 10^-14.

Thus:

K_b = 1.0 × 10^-14 / 1.78 × 10^-4 ≈ 5.62 × 10^-11

To find the pH, we'll solve for the hydrolysis of the conjugate base:

[OH^-] = √(K_b × [HCOO^-]) = √(5.62 × 10^-11 × 0.25)

[OH^-] ≈ √(1.405 × 10^-11) ≈ 3.75 × 10^-6

Then, pOH = -log[OH^-] ≈ -log(3.75 × 10^-6) ≈ 5.43

Finally, pH = 14 - pOH ≈ 14 - 5.43 ≈ 8.57

Conclusion and Indicator Selection

The pH at the equivalence point of the titration of formic acid with sodium hydroxide is approximately 8.6, indicating a slightly basic solution due to the formation of the conjugate base. For such titrations, an appropriate indicator should change color in the basic pH range near this value.

Phenolphthalein, which transitions from colorless in acidic conditions to pink in basic conditions around pH 8.2 to 10.0, is an ideal choice for this titration because it provides a clear end-point close to the equivalence point pH.

In summary, at the equivalence point, the solution contains primarily the conjugate base of formic acid, and the pH is approximately 8.6. Choosing phenolphthalein as an indicator ensures an accurate visual endpoint for this titration.

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