A 25.00 mL sample of a diluted oxalix acid solution will be titrated with NaOH.

Question

A 25.00 mL sample of a diluted oxalix acid solution will be titrated with NaOH. You want to use approximately 20.00 mL of 0.10 M NaOH to copmpletely neutralize the oxalic acid in that sample. What is the concentration of oxalic acid in this diluted oxalic acid solution?

You will prepare the diluted oxalic acid solution from the saturdated oxalic acid solution, which has a concentration of approximately 1.0M. What volume of saturated oxalic acid solution must be used to make 100.00 mL of a diluted oxalic acid solution with the concentration determined in Step 1?

Explanation / Answer

Step1 Moles of NaOH = 20.0x10^-3 x.1 = .002=2.0x10&-3

Step2 Moles of OxalicAcid = (1/2) x Moles of NaOH =.001=1.0x10^-3

Step3 Concentration of Oxalic acid = 1.0x10^-3/.025 = 4.0x10^-2 Mole/litre

Step4 Use M1V1 =M2V2

Step5 1.0xV1 = 4.0x10^-2 x .1 or V1 = 4.0x10^-3 l =4.0ml

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