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Calculation of NaOH Concentration from Neutralization of Oxalic Acid Solution

This document presents a step-by-step calculation to determine the concentration of a sodium hydroxide (NaOH) solution after neutralizing a given volume of oxalic acid (C2H2O4). Initially, 2.35 g of oxalic acid was dissolved in 250 cm³ of water, and a portion of this solution (19.6 cm³) was neutralized by 25 cm³ of NaOH. Using stoichiometric relationships from the balanced chemical equation, the moles of oxalic acid and NaOH are calculated, leading to the determination of the NaOH concentration in mol/dm³.

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Calculation of NaOH Concentration from Neutralization of Oxalic Acid Solution

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  1. Starter am w/c 03-Dec-12 • 2.35 g oxalic acid (C2H2O4) was dissolved into 250 cm3 water. 19.6 cm3 of this solution was found to be neutralised by 25 cm3NaOH. Calculate the concentration of the NaOH solution. The reaction is given below: C2H2O4 + 2NaOH → C2O4Na2 + 2H2O

  2. Starter am w/c 03-Dec-12 • 2.35 g oxalic acid (C2H2O4) was dissolved into 250 cm3 water. 19.6 cm3 of this solution was found to be neutralised by 25 cm3NaOH. Calculate the concentration of the NaOH solution. moles oxalic acid in 250 cm3 = 2.35/90 = 0.0261 moles oxalic acid in 19.6 cm3 = 0.0261 x 19.6/250 = 0.00205 moles NaOH = 0.00205 x 2 = 0.00409 concNaOH = 0.00409 x 1000/25 = 0.164 moldm-3 (3 s.f.)

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