fresh solution of sodium hydroxide. She carefuly weighs out 118, mg of axalic ac

Question

fresh solution of sodium hydroxide. She carefuly weighs out 118, mg of axalic acd (H,C,o) a diprotic acid in high purity, and dissolves it in 250. mL of distilled water. The student then titrates the oxalic acid solution with her A chemistry student needs to sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 70.6 mL of sodium hydroxide solution. Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 3 significant digits

Explanation / Answer

Mass of oxalic acid used = 118 mg =0.118 g

Molar mass of oxalic acid = 90.03 g/mol

Moles of oxalic acid = 0.118 g/ (90.03 g/mol)= 0.0013 mol

This means the student has titrated oxalic acid solution having 0.0013 mol of it.

Now, the reaction of reaction of oxalic acid with NaOH can be written as:

HOOC-COOH + 2 NaOH ---------> NaOOC-COONa + 2 H2O

This means,

1 mol oxalic acid react with NaOH = 2 mol

0.0013 mol oxalic acid react with NaOH =.0013×2=0.0026 mol

Molarity of NaOH solution = moles of NaOH/ volume used in L

= (0.0026 mol)/0.0706 L

= 0.037 mol/L = 0.037 M

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