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iren [92.7K]
2 years ago
5

During the titration of a diluted vinegar sample with a sodium hydroxide solution, the volume of sodium hydroxide used was less

than expected. Which of the following could account for the lower than expected volume?
A. The sodium hydroxide solution had been allowed to stand exposed to the air for a long time prior to the titration.
B. The volumetric flask used to prepare the diluted vinegar solution was rinsed with water prior to use.
C. The burette used to deliver the sodium hydroxide solution was rinsed with water prior to use.
D. The pipette used to deliver the vinegar solution was rinsed with water prior to use.
Chemistry
1 answer:
lapo4ka [179]2 years ago
6 0

Answer:

B and D could be true

Explanation:

A volume of sodium hydroxide less than expected could occurs for two reasons:

The real concentration of sodium hydroxide was higher than expected or the amount of vinegar added was less than expected:

A. The sodium hydroxide solution had been allowed to stand exposed to the air for a long time prior to the titration.  FALSE. A long expose  to the air decreases concentration of the NaOH.

B. The volumetric flask used to prepare the diluted vinegar solution was rinsed with water prior to use.  TRUE. You add a less amount of vinegar doing you require less amount of NaOH than expected.

C. The burette used to deliver the sodium hydroxide solution was rinsed with water prior to use.  FALSE. Thus, you add a less amount of NaOH than expected. To explain the matter, you add more NaOH than expected.

D. The pipette used to deliver the vinegar solution was rinsed with water prior to use. TRUE. Again, you are adding a less amount of Vinegar than expected doing the necessary NaOH during titration less than expected

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<u>Answer:</u>

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The given chemical reaction follows:

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The expression of K_p for the above reaction follows:

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We are given:

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Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

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K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

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R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

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Putting values in above equation, we get:

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