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?
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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