JEE Challenger
More from Some Basic Concepts of Chemistry
ChemMedAdv 2020 • Paper 1

Calculate Molarity of Sodium Hydroxide Solution from Titration Data

A 5.00 mL5.00\text{ mL} volume of 0.10 M0.10\text{ M} oxalic acid solution placed in a conical flask is titrated against a NaOH\text{NaOH} solution from a burette using phenolphthalein indicator. The volume of NaOH\text{NaOH} required to observe a faint permanent pink color end point is listed below for five experimental runs. Calculate the molar concentration of the NaOH\text{NaOH} solution.

Exp. No.Vol. of NaOH (mL)112.5210.539.049.059.0\begin{array}{|c|c|} \hline \text{Exp. No.} & \text{Vol. of NaOH (mL)} \\ \hline 1 & 12.5 \\ 2 & 10.5 \\ 3 & 9.0 \\ 4 & 9.0 \\ 5 & 9.0 \\ \hline \end{array}
Official Numerical Answer0.11

Topics & Concepts

Some Basic Concepts of ChemistryConcentration TermsStoichiometry

Step-by-Step Solution

To calculate the molarity of the NaOH\text{NaOH} solution, we identify the concordant volume of NaOH\text{NaOH} used in the titration from the three identical consecutive runs, which is VNaOH=9.0 mLV_{\text{NaOH}} = 9.0\text{ mL}.

Oxalic acid (H2C2O4\text{H}_2\text{C}_2\text{O}_4) is a dibasic acid (n=2n = 2), while NaOH\text{NaOH} is a monoacidic base (n=1n = 1). At the equivalence point, the milliequivalents of NaOH\text{NaOH} must equal the milliequivalents of oxalic acid:

MNaOH×VNaOH×1=Moxalic×Voxalic×2M_{\text{NaOH}} \times V_{\text{NaOH}} \times 1 = M_{\text{oxalic}} \times V_{\text{oxalic}} \times 2

Substituting the given values into the equation:

MNaOH×9.0 mL=0.10 M×5.00 mL×2M_{\text{NaOH}} \times 9.0\text{ mL} = 0.10\text{ M} \times 5.00\text{ mL} \times 2

MNaOH=1.09.0≈0.11 MM_{\text{NaOH}} = \frac{1.0}{9.0} \approx 0.11\text{ M}

Thus, the molar concentration of the NaOH\text{NaOH} solution is 0.11 M0.11\text{ M} (or 0.111 M0.111\text{ M}).

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