Question

Oxalic acid is a common dibasic acid. Draw a qualitative sketch of the titration curve for a sample of oxalic acid titrated with a solution of NaOH. Label the end point(s). Assume the pKa's of oxalic acid are approximately 1.2 and 4.2

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Answer 1

Let's consider 25 mL of 0.1 M oxalic acid, which you are titrating with 0.1 M NaOH.

The no. of mmol of oxalic acid (n_{oxalic acid}) = 25 mL * 0.1 mmol/mL = 2.5 mmol

(i) Initial pH = 1/2 (pKa1 - Log[oxalic acid])

i.e. pH = 1/2 (1.2 - Log0.1)

i.e. pH = 1.1

(ii) Volume of 0.1 M NaOH = 5 mL

i.e. The no. of mmol of NaOH (n_NaOH) = 5 mL * 0.1 mmol/mL = 0.5 mmol

According to Henderson-Hasselbulch equation:

pH = pKa1 + Log{n_NaOH/(n_{oxalic acid} - n_NaOH)}

i.e. pH = 1.2 + Log{0.5/(2.5-0.5)}

i.e. pH = 0.598

(iii) Volume of 0.1 M NaOH = 10 mL

i.e. The no. of mmol of NaOH = 10 mL * 0.1 mmol/mL = 1 mmol

According to Henderson-Hasselbulch equation:

pH = pKa1 + Log{n_NaOH/(n_{oxalic acid} - n_NaOH)}

i.e. pH = 1.2 + Log{1/(2.5-1)}

i.e. pH = 1.024

(iv) Volume of 0.1 M NaOH = 12.5 mL

i.e. The no. of mmol of NaOH = 12.5 mL * 0.1 mmol/mL = 1.25 mmol

According to Henderson-Hasselbulch equation:

pH = pKa1 + Log{n_NaOH/(n_{oxalic acid} - n_NaOH)}

i.e. pH = 1.2 + Log{1.25/(2.5-1.25)}

i.e. pH = 1.2 (first equivalence point)

(v) Volume of 0.1 M NaOH = 20 mL

i.e. The no. of mmol of NaOH = 20 mL * 0.1 mmol/mL = 2 mmol

According to Henderson-Hasselbulch equation:

pH = pKa1 + Log{n_NaOH/(n_{oxalic acid} - n_NaOH)}

i.e. pH = 1.2 + Log{2/(2.5-2)}

i.e. pH = 1.802

(v) Volume of 0.1 M NaOH = 25 mL

i.e. The no. of mmol of NaOH = 25 mL * 0.1 mmol/mL = 2.5 mmol

Here, [monooxalate] = 2.5 mmol/(25 + 25) mL = 0.05 M

Now, pH = 7 + 1/2 (pKa1 + Log[monooxalate])

= 7 + 1/2 (1.2 + Log0.05)

pH = 6.95 (first equivalence point)

(vi) Volume of 0.1 M NaOH = 30 mL

i.e. The no. of mmol of NaOH = 30 mL * 0.1 mmol/mL = 3 mmol

n_monooxalate = 2.5 mmol

n_NaOH = 3-2.5 = 0.5 mmol

According to Henderson-Hasselbulch equation:

pH = pKa2 + Log{n_NaOH/(n_monooxalate - n_NaOH)}

i.e. pH = 4.2 + Log{0.5/(2.5-0.5)}

i.e. pH = 3.598

(vii) Volume of 0.1 M NaOH = 35 mL

i.e. The no. of mmol of NaOH = 35 mL * 0.1 mmol/mL = 3.5 mmol

n_monooxalate = 2.5 mmol

n_NaOH = 3.5-2.5 = 1 mmol

According to Henderson-Hasselbulch equation:

pH = pKa2 + Log{n_NaOH/(n_monooxalate - n_NaOH)}

i.e. pH = 4.2 + Log{1/(2.5-1)}

i.e. pH = 4.024

(viii) Volume of 0.1 M NaOH = 37.5 mL

i.e. The no. of mmol of NaOH = 37.5 mL * 0.1 mmol/mL = 3.75 mmol

n_monooxalate = 2.5 mmol

n_NaOH = 3.75-2.5 = 1.25 mmol

According to Henderson-Hasselbulch equation:

pH = pKa2 + Log{n_NaOH/(n_monooxalate - n_NaOH)}

i.e. pH = 4.2 + Log{1.25/(2.5-1.25)}

i.e. pH = 4.2 (second half-equivalence point)

(ix) Volume of 0.1 M NaOH = 45 mL

i.e. The no. of mmol of NaOH = 45 mL * 0.1 mmol/mL = 4.5 mmol

n_monooxalate = 2.5 mmol

n_NaOH = 4.5-2.5 = 2.0 mmol

According to Henderson-Hasselbulch equation:

pH = pKa2 + Log{n_NaOH/(n_monooxalate - n_NaOH)}

i.e. pH = 4.2 + Log{2/(2.5-2)}

i.e. pH = 4.802

(x) Volume of 0.1 M NaOH = 50 mL

i.e. The no. of mmol of NaOH = 50 mL * 0.1 mmol/mL = 5 mmol

Here, [dioxalate] = 2.5 mmol/(25 + 50) mL = 0.033 M

Now, pH = 7 + 1/2 (pKa2 + Log[dioxalate])

= 7 + 1/2 (4.2 + Log0.033)

pH = 8.36 (second equivalence point)

Now, the plot of pH versus volume of NaOH added can be drawn as follows.