Question

As an Analytical chemist, you are designing a titration experiment to determine formic acid in food samples (formic acid is used as a preservative and antibacterial agent). You test your protocol by performing a titration of 100.0 mL of 0.050 M formic acid (FW = 46.03 g/mol) with a 0.10 M sodium hydroxide standard solution. For formic acid K_a = 1.80 x 10⁻⁴ You can ignore activity coefficients!    (60 points total) V_b = Volume of base added NOTE: Just providing the pH-values will not get you full credit. You must provide your work! a. Determine the volume (mL) of base needed to reach the equivalence point. (4 points) b. Calculate the pH corresponding to V_b = 0.0 mL (10 points) Answer: pH = __________​​​​​​​ c. Calculate the pH corresponding to V_b = 15.0 mL (9 points) Answer: pH = __________​​​​​​​ d. Calculate the pH corresponding to V_b = 25.0 mL (9 points) Answer: pH = __________​​​​​​​ e. Calculate the pH corresponding to V_b = 45.0 mL (9 points) Answer: pH = __________​​​​​​​ f. Calculate the pH corresponding to V_b = 50.0 mL (10 points) Answer: pH = __________​​​​​​​ g. Calculate the pH corresponding to V_b = 55.0 mL (9 points) Answer: pH = __________ h. For the titration performed, you use the indicator cresol purple (see the Table include – Table 11-3). Describe what color you would see at EACH of the points above (i.e. b – g) for the titration! (6 points)

Answer 1

Before the addition of the base. The hydrolysis of the acid is taken into account:

HCHO2 + H2O = CHO2- + H3O+

We have the expression for Ka:

Ka = [CHO2-] * [H3O+] / [HCHO2]

It is replaced:

0.00018 = X ^ 2 / 0.05 - X

Clears:

X ^ 2 + 0.00018 * X - 9e-06 = 0

The second degree equation is applied and we have:

X = 0.0029 M = [H3O+]

The pH is calculated:

pH = - log [H3O+] = - log 0.0029 = 2.54

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Before the equivalence point. The initial HCHO2, added base, and resulting HCHO2 moles are calculated:

n Initial HCHO2 = M * V = 0.05 M * 0.1 L = 0.005 mol

n NaOH = 0.1 M * 0.015 L = 0.0015 mol = n CHO2-

n Final HCHO2 = n Initial HCHO2 - n NaOH = 0.005 - 0.0015 = 0.0035 mol

Using the Henderson-Hasselbach equation, the pH is calculated:

pH = pKa + log (n NaOH / n final HCHO2) = 3.74 + log (0.0015 / 0.0035) = 3.38

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Before the equivalence point. The initial HCHO2, added base, and resulting HCHO2 moles are calculated:

n Initial HCHO2 = M * V = 0.05 M * 0.1 L = 0.005 mol

n NaOH = 0.1 M * 0.025 L = 0.0025 mol = n CHO2-

n Final HCHO2 = n Initial HCHO2 - n NaOH = 0.005 - 0.0025 = 0.0025 mol

Using the Henderson-Hasselbach equation, the pH is calculated:

pH = pKa + log (n NaOH / n final HCHO2) = 3.74 + log (0.0025 / 0.0025) = 3.74

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Before the equivalence point. The initial HCHO2, added base, and resulting HCHO2 moles are calculated:

n Initial HCHO2 = M * V = 0.05 M * 0.1 L = 0.005 mol

n NaOH = 0.1 M * 0.045 L = 0.0045 mol = n CHO2-

n Final HCHO2 = n Initial HCHO2 - n NaOH = 0.005 - 0.0045 = 0.0005 mol

Using the Henderson-Hasselbach equation, the pH is calculated:

pH = pKa + log (n NaOH / n final HCHO2) = 3.74 + log (0.0045 / 0.0005) = 4.7

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At the equivalence point. The hydrolysis reaction of the conjugated base is taken into account:

CHO2- + H2O = HCHO2 + OH-

The total volume of solution is calculated at the equivalence point:

V NaOH = Ca * Va / Cb = 0.05 M * 100mL / 0.1 M = 50 mL

Total V = V HCHO2 + V NaOH = 100 + 50 = 150 mL

The initial CHO2- concentration at the equivalence point is calculated for a total volume of 150 mL:

Initial [CHO2-] = C1 * V1 / V2 = 0.05 M * 100 mL / 150 mL = 0.033 M

We have the expression of Kb:

Kb = [HCHO2] * [OH-] / [CHO2-]

5.6e-11= X^2 / 0.033 - X

Clears:

X^2 + 5.6e-11 * X - 1.9e-12 = 0

The second degree equation is applied and we have:

X = 1.4e-06 M = [OH-]

The pOH and pH are calculated:

pOH = - log [OH-] = - log 1.4e-06 = 5.87

pH = 14 - pOH = 8.13

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After the equivalence point. The initial acid moles, added base and remaining base are calculated:

n Initial HCHO2 = M * V = 0.05 M * 0.1 L = 0.005 mol

n added NaOH = 0.1 M * 0.055 L = 0.0055 mol

n remaining NaOH = n added NaOH - n initial HCHO2 = 0.0055 - 0.005 = 0.0005 mol

We have the total volume:

Total V = 100 mL + 55 mL = 155 mL

The remaining NaOH concentration in the total volume is calculated:

[NaOH] = n / V = 0.0005 / 0.15 = 0.0032 M

The pOH and pH are calculated:

pOH = - log 0.0032 = 2.49

pH = 14 - pOH = 11.51

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