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Practical Organic Chemistry question

2021 · 25 Feb · Shift 2 · Q19
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Practical Organic Chemistry question

2021 · 25 Feb · Shift 2 · Q19

JEE MainChemistryPractical Organic ChemistryNumerical+4 / −1
Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained. (i) 4.5 mL (ii) 4.5 mL (iii) 4.4 mL (iv) 4.4 mL (v) 4.4 mL If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is ‾\underline{\hspace{2cm}}​ M. (Rounded off to the nearest integer)
Numerical answer
View written solutionFree

Correct answer: 6

  1. Given data
  • Molarity of oxalic acid =1.25 M= 1.25\,\text{M}=1.25M
  • Volume of oxalic acid taken =10.0 mL= 10.0\,\text{mL}=10.0mL
  • Burette readings for NaOH used: 4.5, 4.5, 4.4, 4.4, 4.4 mL4.5,\,4.5,\,4.4,\,4.4,\,4.4\,\text{mL}4.5,4.5,4.4,4.4,4.4mL
  1. Find the concordant/average burette reading

Since the readings are very close, take the average:

VNaOH=4.5+4.5+4.4+4.4+4.45=22.25=4.44 mLV_{\text{NaOH}} = \frac{4.5+4.5+4.4+4.4+4.4}{5} = \frac{22.2}{5} = 4.44\,\text{mL}VNaOH​=54.5+4.5+4.4+4.4+4.4​=522.2​=4.44mL
  1. Write the reaction

Oxalic acid is dibasic:

H2C2O4+2NaOH→Na2C2O4+2H2O\mathrm{H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O}H2​C2​O4​+2NaOH→Na2​C2​O4​+2H2​O

So,

1 mole oxalic acid:2 moles NaOH1 \text{ mole oxalic acid} : 2 \text{ moles NaOH}1 mole oxalic acid:2 moles NaOH
  1. Moles of oxalic acid used
nacid=M×V=1.25×10.01000n_{\text{acid}} = M \times V = 1.25 \times \frac{10.0}{1000}nacid​=M×V=1.25×100010.0​ nacid=0.0125 moln_{\text{acid}} = 0.0125\,\text{mol}nacid​=0.0125mol
  1. Moles of NaOH required

Using the stoichiometric ratio 1:21:21:2,

nNaOH=2×0.0125=0.025 moln_{\text{NaOH}} = 2 \times 0.0125 = 0.025\,\text{mol}nNaOH​=2×0.0125=0.025mol
  1. Calculate molarity of NaOH

Volume of NaOH used =4.44 mL=0.00444 L= 4.44\,\text{mL} = 0.00444\,\text{L}=4.44mL=0.00444L

MNaOH=nV=0.0250.00444M_{\text{NaOH}} = \frac{n}{V} = \frac{0.025}{0.00444}MNaOH​=Vn​=0.004440.025​ MNaOH≈5.63 MM_{\text{NaOH}} \approx 5.63\,\text{M}MNaOH​≈5.63M
  1. Round off to nearest integer
5.63≈65.63 \approx 65.63≈6

Final Answer

The molarity of NaOH solution is:

6\boxed{6}6​
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