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A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out of oxalic acid , a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in of distilled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used of sodium hydroxide solution.Calculate the molarity of the student's sodium hydroxide solution. Be sure your answer has the correct number of significant digits.

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Answer:

See explanation below to full answer

Explanation:

First of all, you are not providing the amounts of acid and hydroxide here, to do the calculations. However, in order to help you, I will use these values that are taken from a similar exercise. Then, replace your data with this procedure and you should get the correct answer.

For this part, I will say that the student weights about 210 mg of oxalic acid, (H2C2O4) and the volume of NaOH used to reach equivalent point was 150 mL in a beaker of 250 mL.

Now the equivalence point is the point where both moles of acid and hydroxide are the same. In other words:

nA = nB

The reaction that it's taking place is the following:

2NaOH + H2C2O4 ----------> Na2C2O4 + 2H2O

This means that 2 moles of NaOH reacts with 1 mole of H2C2O4, therefore the expression in (1) corrected is:

nB = 2 nA

So, we need to calculate first the moles of the acid. To do that we need the molar mass of the acid (the reported is 90.03 g/mol)

nA = 0.210 / 90.03 = 0.0023 moles

We have the moles of acid used, so the moles of the hydroxide is:

nB = 2 * 0.0023 = 0.0046 moles

We have the volume used of hydroxide, which is 150 mL, so finally the concentration is:

MB = 0.0046 / 0.150 = 0.031 M

Now, replace the actual values that you have in here, and you should get an accurate result.