Suppose a student failed to dry (remove the water from) the KHP before using it

Question

Suppose a student failed to dry (remove the water from) the KHP before using it to standardize the NaOH solution. What effect would this have on the calculated molarity of the NaOH solution?

Question 6 options:

The calculated molarity would be too low.

The calculated molarity would be too high.

It would have no effect on the calculated molarity.

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Question 7 (2 points)

Suppose a student failed to dry (remove the water from) the KHP before using it to standardize the NaOH solution. What effect would this have on the calculated equivalent mass of the unknown acid?

Question 7 options:

The calculated equivalent mass would be too low.

The calculated equivalent mass would be too high.

It would have no effect on the calculated equivalent mass.

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Question 8 (2 points)

Suppose a student performing this experiment calibrated the pH meter using the pH 6 buffer instead of the pH 7 buffer. As a result, all the pH meter readings were too low. What effect would this procedural error have on the experimentally determined pKa of the unknown acid?

Question 8 options:

The pKa would be too low.

The pKa would not be affected.

The pKa would be too high.

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The calculated molarity would be too low.

The calculated molarity would be too high.

It would have no effect on the calculated molarity.

Explanation / Answer

ANSWER

Question 6. Option (B) is correct answer. If KHP is impue (for example containing water) more mass of KHP will be required to furnish the required moles of H+ for neutralization of NaOH. Hence more will be the calculated molarity of NaOH.

Question 7. Option (A) is correct answer. As the calculated concentration of NaOH is high due to impure KHP, the amount of H+ to neutralize OH- from NaOH will also be high, so equivalent mass will be low as is evident from the following equation:

equivalent mass = mass of acid / moles of H+ titrated.

moles of H+ titrated will be higher hence equivalent mass is lower.

Question 8: Option (A) is correct answer.

Consider an acid HA

HA <----> H+ + A-

As all the pH meter reading are too low, the H+ concentrations will be too high, because pH is the negative logrithm of H+ concentration. As H+ is too high the Ka value will also be too high as per the equation

Ka = [H+][A-] / [HA]

A higher Ka value means lower pKa value , because pKa is the negative logrithm of Ka.

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