When standardizing the NaOH with KHP, if the KHP was overtitrated, How would thi
ID: 1017199 • Letter: W
Question
When standardizing the NaOH with KHP, if the KHP was overtitrated, How would this affect the molarity of NaOH? How would this affect the molarity of the acid unknowns? How would the molarity of the acids and their calculated K_a values (where applicable) be affected if too much phenolphthalein (an acid) was added? Could this experiment be performed without indicators? Explain. When titrating the acid solutions, if the beaker was rinsed with DI water, but not dried, How would this affect the initial pH? How would this affect the NaOH mL to the equivalence point and the pH of the equivalence point? For a weak base - strong acid titration, describe two methods that can be used to determine K_b (K_w = K_a K_b is not a valid answer for this question.)Explanation / Answer
Error effect on experiment
1. Standarization of NaOH with KHP : overtitrated
a. The calculate molarity of NaOH would be higher as the moles of KHP would be higher due to volume KHP added is greater than actual volume (moles = molarity x volume).
b. molarity of acid thus calculated would be higher than actual value.
2. If phenolphthalein added was excess, the molarity of acid would remain unaffected. Thus Ka value would also remain just the same.
3. No the experiment cannot be done in the absence of indicator. The end point would not be identified easily and we would get incorrect results.
4. If beaker was rinsed with DI water, but not dired.
a. The overall volume of solution to be titrated would increase. pH would be higher. More dilute solution means lower concentration and greater dissociation.
b. No effect on NaOH volume would be seen as moles of acid remains the same in solution. pH at equivalence point would thus also remain the same.
5. For weak base-strong acid titration.
pH at equivalence point correesponds to [H+]
From it [OH-] concentration can be calculated and that can be fed to Kb = [OH-][BH+]/[B] equation to find Kb
Alternately, the pH at 1/2 Eq point = pKa for conjugate acid of weak base.
pKb = 14 - pKa = -log[Kb]
Thus, Kb can easily be calculated from it.
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