Question

A 0.6972 g of impure potassium hydrogen phthalate (FW: 204.23) was dissolved in 30 mL of water and 2 drops of phenolphthalein was added. It required 24.83 mL of 0.1097 M NaOH for complete titration.

a. how many moles of acid are present?

b. how would the titration be affected if you used 0.8786 g of KHP?

c. if you added 27.75 mL of NaOH instead of 25.83 mL, what would you expect to see? Note, this question does not involve a calculation, rather it is referring to your qualitative observations of the experiment.

d. when setting up their experiment, a student discovered that the solution in their flask turned bright pink upon adding 2 drops of the indicator, even before beginning the titration. what do you suspect was the mistake the student made?

Answer 1

a.

Since it is acid base titration and Naoh is base , KHP is considered as acid.

As KHP is acid ,

no. Of moles of acid = no. Of moles of KHP

No. Of moles of KHP= weight of KHP taken / formula weight

= 0.6972/204.23

= 0.0034 moles .

b.

If 0.8786 g of KHP used then end point will be more instead of 24.83. because larger amount of KHP taken will require more amount of NaOH for neutralization of reaction.

c.

if27.75 ml of NaOH is added then solution will become more basic because required amount is 24.83 and beyond that if excess of NaOH is added solution will turn dark pink in colour .

d.

If you get immediate pink colour after addition of indicator propably the mistake is that student might have taken NaOH in conical flask instead of KHP.

Because phenolphthalein indicates pink colour in basic medium. It do not show pink colour in acidic medium.

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