Question

1. Answer the following question based on the reaction below: NaOH(aq) + KHP(s) ---> NaKp(aq) +...

1. Answer the following question based on the reaction below:

NaOH(aq) + KHP(s) ---> NaKp(aq) + H2O(l)

A 1.348g sample of impure KHP was titrated with a 0.0942 M solution of NaOh. To completely react the KHP in the sample, 69.34mL of base was needed. KHP (potassium hydrogen phthalate, 204.23 g/mol)

a.)How many grams of KHP were in the unknown sample?

b.) What is the precentage of KHP in the unknown sample?

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Answer #1

NaOH(aq) + KHP(s) ---> NaKp(aq) + H2O(l)

no of moles of NaOH = molarity * volume in L

                                    = 0.0942*0.06934   = 0.00653 moles

1 mmoles of NaOH react with 1 mole of KHP

0.00653 moles of NaOH react with 0.00653 moles of KHP

mass of KHP   = no of moles * gram molar mass

                       = 0.00653*204.23   = 1.334g of KHP

b.

The percentage of KHP in unknown sample   = 1.334*100/1.348   = 98.96%

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