Consider titration of NaOH solution versus 1.25M 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 volu

Question

Consider titration of NaOH solution versus 1.25M 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 ______ M. (Rounded-off the nearest integer)

Prashant Kotian · Accepted Answer

Correct answer: 6.00. **Step 1: Identify concordant readings** The burette readings are: 4.5, 4.5, 4.4, 4.4, 4.4 mL. The concordant (consistent) readings are 4.4 mL, appearing three times. Discard 4.5 mL readings as outliers. **Step 2: Calculate volume of NaOH used** Average of concordant readings = (4.4 + 4.4 + 4.4)/3 = 4.4 mL Volume of NaOH used = 4.4 mL **Step 3: Write the balanced equation** Oxalic acid (H₂C₂O₄) is diprotic; NaOH is monobasic: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O Mole ratio: 1 mole of oxalic acid requires 2 moles of NaOH **Step 4: Apply the titration formula** M₁V₁/n₁ = M₂V₂/n₂ Where: - M₁ = 1.25 M (oxalic acid) - V₁ = 10.0 mL (oxalic acid) - n₁ = 2 (number of H⁺ from oxalic acid) - M₂ = ? (NaOH molarity) - V₂ = 4.4 mL (NaOH) - n₂ = 1 (number of OH⁻ from NaOH) **Step 5: Solve for M₂** (1.25 × 10.0)/2 = M₂ × 4.4/1 M₂ = (1.25 × 10.0 × 1)/(2 × 4.4) = 12.5/2.08 = 6.01 M ≈ 6.00 M Therefore

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