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Margaret [11]
3 years ago
15

A 0.500-gram sample of a weak, nonvolatile monoprotic acid, HA, was dissolved in sufficient water to make 50.0 milliliters of so

lution. The solution was then titrated with a standard NaOH solution. Predict how the calculated molar mass of HA would be affected (too high, too low, or not affected by the following laboratory procedures. Explain each of your answers.
a. After rinsing the buret with distilled water, the buret is filled with the standard NaOH solution; the weak acid HA is titrated to its equivalence point.
b. Extra water is added to the 0.500-gram sample of HA.
(a) An indicator that changes color at pH 5 is used to signal the equivalence point.
(b) An air bubble passes unnoticed through the tip of the buret during the titration.
Chemistry
1 answer:
hodyreva [135]3 years ago
6 0

Answer:

Follows are the solution to the given points:

Explanation:

  • It is expected to be too low a calculated molecular weight. It is because NaOH is diluted throughout washing with distilled water. It means that its measured intensity would be too high. Mwt implies Mwt now. Oh, Mwt. It's low.
  • If the water becomes inserted throughout the acid HA, the molarity Across would not be impacted as the acidic mole remains unchanged despite water added to the acidic Ha.
  • It's going to be much too big a molecular weight. Its measurement indicates the middle state at PH=5, well before this neutral point, so that if the NaOH reading is less, and the value of the strength of HA would consequently become lower. So m/Mwt=low shows Mwt=high.
  • Its real measurement will be less because of the air pocket in the office. Even so, because of an unknown bubble interpretation, more is removed. Its intensity of acid is, therefore, less measured, and thus Mwt is also too high.
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<u>____________________________</u>

Explanation:

<u>____________________________</u>

<u>Note</u>: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

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The "balanced chemical equation" is:

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        2 Al   +   3 Cl₂   →   2 AlCl₃   ;

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<u>Note</u>: The molecular weight of "aluminum (Al)" is:   " 26.98 g /mol " .

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So:  We call solve using a technique known as:  "dimensional analysis" :

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  0.500 mol AlCl₃ * (\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?

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<u>Note</u>:  The units of "mol AlCl₃" cancel out to "1' ; and:

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 " \frac{(0.500 * 2 * 26.98)}{2}   g Al ["grams of aluminum"] ;

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<u>Note</u>: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

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         →  Round to 3 (Three) significant figures;

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 Hope this is helpful!  

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