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Lelechka [254]
1 year ago
12

You prepare a standard by weighing 10.751 mg of compound X into a 100 mL volumetric flask and making to volume. You further dilu

te this solution 5 mL to 25 mL. This standard gives an area of 4,374. Your sample is prepared by adding 5 mL of sample solution into a 50 mL flask and making to volume. This gives an area count of 2,582. Calculate the concentration of compound X in the sample - prior to dilution.
Chemistry
1 answer:
ArbitrLikvidat [17]1 year ago
4 0

Answer:

0.12693 mg/L

Explanation:

First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:

  • 10.751 mg / 100 mL = 0.10751 mg/mL

Then we <u>calculate the concentration of compound X in the standard after dilution</u>:

  • 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L

Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:

  • 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L

Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:

  • 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L
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So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

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The reduction of a chemical species is represented in a reduction half- reaction and the oxidation of a chemical species is represented in a oxidation half- reaction.

<u>To balance the reduction half-reaction for the reduction of Cr₂O₇²⁻ to Cr³⁺</u>:

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