In: Chemistry
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?
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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?
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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?
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1) If a student failed to dry (remove the water from) the KHP before using it to standardize the NaOH solution. The resultant molarity calculated of the NaOH solution would be,
B) The calculated molarity would be too low.
1 mole of KHP reacts with 1 mole of NaOH. Since the actual concentration of KHP would be low by increasng the initial weight due to presence of water in it, the net molar concentration of NaOH would also be low.
2) If a student failed to dry (remove the water from) the KHP before using it to standardize the NaOH solution. The effect on the calculated equivalent mass of the unknown acid would be,
B) the calculated equivalent mass of acid would be too high
moles of base calculated would be low here, which automatically means moles of acid neutralized would be high as calculate. So the equivalent mass would also be high.
3) If 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. The calculated pKa of the unknown acid would thus turn out to be,
C) The pKa would be too low
at half equivalence point pH measured would be lower and this implies the pKa to be low too for the acid.