POST-LAB QUESTIONS 1. Prepare a theoretical titration curve for titration of 25.

Question

POST-LAB QUESTIONS 1. Prepare a theoretical titration curve for titration of 25.0 mL of 0.1037 M formic acid (HCOOH: pKa 3.75) solution (diluted to 100 mL volume with deionized water) by 0.0964 M solution of KOH: a. Determine the volume of KOH solution needed to reach the equivalence point b. Calculate pH of the starting solution (remember, it was diluted to 100 mL). c. Calculate pH at the equivalence point.(The total volume is now more than 100 mL!) d. Calculate pH at points between the start of titration and equivalence point; choose points so that they will be more dense where change in pH is faster. A good guess would be to go every 0.1 ml within 1 mL of equivalence point, every 0.5 mL up to 5 mL from equivalence point, and every 2 mL outside of this interval. Excel (or similar program) is very useful for this! Calculate pH at points frorn equivalence point up until 150% of volume at equivalence point. c. Plot the points obtained in Parts (B) (E) into the graph. Did you obtain a smooth curve? Why or why not? f.

Explanation / Answer

a. At the equivalence point, the no. of millimoles of formic acid = no. of millimoles of KOH

i.e. M1V1 = MV2, where 1 = formic acid and 2 = KOH

i.e. 0.1037 M * 25 mL = 0.0964 M * V2

i.e. V2 = (0.1037*25/0.0964) mL = 26.9 mL

b. The no. of mmol of formic acid = 0.1037 mmol/mL * 25 mL = 2.5925 mmol

The no. of mmol of KOH = 0.0964 mmol/mL * 100 mL = 9.64 mmol

Here, 2.5925 mmol of formic acid reacts with 2.5925 mmol of KOH to form 2.5925 mmol of potassium formate. Then the remaining no. of mmol of KOH = 9.64 - 2.5925 = 7.0475 mmol

Now, the [KOH] = [OH-] = 7.0475 mmol/100 mL = 0.070475 M

Now, pOH = -Log[OH-] = -Log(0.070475) = 1.15

pH + pOH = 14

i.e. pH + 1.152 = 14

i.e. pH = 14 - 1.15 = 12.85

c. At the equivalence point, the concentration of potassium formate = 2.5925 mmol/100 mL = 0.025925

Formula: pH = 7 + 1/2 (pKa + Log[potassium formate])

i.e. pH = 7 + 1/2 (3.75 + Log 0.025925)

i.e. pH = 8.08

d. (i) Let's say the volume of KOH = 26 mL (with in 1 mL of equivalence point)

The no. of mmol of KOH = 0.0964 mmol/mL * 26 mL = 2.5064 mmol

Here, 2.5064 mmol of formic acid reacts with 2.5064 mmol of KOH to form 2.5064 mmol of potassium formate. Then the remaining mmol of formic acid = 2.5925 - 2.5064 = 0.0861 mmol

According to Henderson-Hasselbulch equation: pH = pKa + Log(nKOH/nformic acid)

i.e. pH = 3.75 + Log(2.5064/0.0861)

i.e. pH = 5.214

(ii) Let's say the volume of KOH = 24.4 mL (with in 5 mL of equivalence point)

The no. of mmol of KOH = 0.0964 mmol/mL * 24.4 mL = 2.35216 mmol

Here, 2.35216 mmol of formic acid reacts with 2.35216 mmol of KOH to form 2.35216 mmol of potassium formate. Then the remaining mmol of formic acid = 2.5925 - 2.35216 = 0.24034 mmol

According to Henderson-Hasselbulch equation: pH = pKa + Log(nKOH/nformic acid)

i.e. pH = 3.75 + Log(2.35216/0.24034)

i.e. pH = 4.741

(iii) Let's say the volume of KOH = 28.9 mL (2 mL outside the equivalence point)

The no. of mmol of KOH = 0.0964 mmol/mL * 28.9 mL = 2.78596 mmol

Here, 2.5925 mmol of formic acid reacts with 2.5925 mmol of KOH to form 2.5925 mmol of potassium formate. Then the remaining mmol of KOH = 2.78596 - 2.5925 = 0.19346 mmol

The concentration of KOH = 0.19346 mmol/(25 + 28.9) mL = 0.0036 M

i.e. pOH = -Log(0.0036)

i.e. pOH = 2.445

i.e. pH = 14 - 2.44 = 11.555

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