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Oxalic Acid Titration

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Besides, the neutralization balance equation for the titration of oxalic acid (C2H2O4) with sodium hydroxide (NaOH) is as below. C2H2O4 (aq) + 2NaOH (aq) Na2C2O4 (aq) + 2H2O (l)
From the equation, it clearly showed that the 1 mole of oxalic acid (C2H2O4) will react with 2 mole of sodium hydroxide (NaOH) to produce 1 mole of oxalic acid sodium salt (Na2C2O4) and 2 moles of water (H2O). The mole ratio of oxalic acid and sodium hydroxide is 1:2
Thus, to calculate the actual concentration of sodium hydroxide (NaOH), the formula below was used. Nacid x Vacid no of mole of acid Nbase x Vbase no of mole of base
The concentration of oxalic acid used was 0.10 N and volume was 10.00mL …show more content…

In an acidic solution, there is an excess of hydronium, H3O+ ions present in the solution. So, the equilibrium will shift to the left and favor the formation of Hln. Thus, the initial oxalic acid (C2H2O4) and acetic acid (CH3COOH) solution observed were remained as colourless solution before performing acid-base titration. From the result of Table 1, the oxalic acid (C2H2O4) and acetic acid (CH3COOH) for first trial showed a dark pink colour solution after acid-base titration. This was due to excess hydroxide ions (OH-) from sodium hydroxide (NaOH) had been added into oxalic acid (C2H2O4) and acetic acid (CH3COOH) cause the solution turned into dark pink instead of light pink. This also indicated that the solution had been over-titrated with sodium …show more content…

This means that acid-base titration was stopped when the faint permanent pink colouration was formed. When there are enough sodium hydroxide (NaOH) solution being added to react with all the hydronium ions (H3O+), the next drop of sodium hydroxide (NaOH) will provided an excess amount of hydroxide ions, OH- ions and a basic solution was formed. Then, the excess present of hydroxide ions (OH-) will react with the hydronium ions (H3O+) to form water. This will shifts the equilibrium to the right because water is being formed and hydronium ions (H3O+) are being removed. This also indicated that all the hydronium ions (H3O+) and hydroxide ions (OH-) present in the solution had been successfully neutralized and reached the end point of the titration reaction. Thus, excess hydroxide ions (OH-) caused the solution to turn into pink colour and pH at range 8.0-9.0. The net ionic equation for the neutralization reaction between hydronium ions (H3O+) and hydroxide ions (OH-) were showed as

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