The student found that 25.0 cm3 of the sodium hydroxide solution was neutralised by 15.00 cm3 of the 0.0480 mol/dm3 ethanedioic acid solution. The equation for the reaction is: H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O Calculate the concentration of the sodium hydroxide solution in mol/dm3 – 6137

One thought on “The student found that 25.0 cm3 of the sodium hydroxide solution was neutralised by 15.00 cm3 of the 0.0480 mol/dm3 ethanedioic acid solution. The equation for the reaction is: H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O Calculate the concentration of the sodium hydroxide solution in mol/dm3 – 6137”

1. Step 1:
   Moles of oxalic acid (H₂C₂O₄) =
   15.0
   1000
   ×
   0.0480
   =
   0.00072
   1000
   15.0
   ​
   ×0.0480=0.00072 mol

   Step 2:
   Moles of NaOH = moles of H₂C₂O₄ × 2
   =
   0.00072
   ×
   2
   =
   0.00144
   0.00072×2=0.00144 mol

   Step 3:
   Concentration of NaOH =
   0.00144
   25.0
   ×
   1000
   =
   0.0576
   25.0
   0.00144
   ​
   ×1000=0.0576 mol/dm³

   Final Answer:
   Concentration = 0.0576 mol/dm³ (allow 0.058 mol/dm³)

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