KHP titration help
answer the following questions based on the reaction below: NaOH(aq) + KHP(s) --> NaKP(aq)+H2O(I)
A 1.864 g sample of impure KHP was titrated with a 0.0978 M solution of NaOH. To completely react the KHP in the sample, 72.14 mL 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 percentage of KHP in the unknown sample?
by 
Perkulator
ANSWERS: 1
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1 mole of NaOH reacts with 1 mole of KHP. Find the moles of NaOh in the reaction to get the moles of KHP, since the ratio is 1:1. NaOH required in the titration is 72.14 mL of 0.0978 mol/L. Multiply the L times mol/L to get moles. 72.14 mL = 72.14 mL / 1000 mL/L = 0.07214 L. 0.07214 L x 0.0978 mol/L = 0.007055 moles. 0.007055 moles NaOH means 0.007055 moles of KHP (if no side reactions interfere, and the titration was done properly). Assuming 0.007055 mol KHP, we convert into g using the conversion given of 204.23 g/mol KHP. 204.23 g/mol x 0.007055 mol = 1.441 g KHP. To get the percent purity, divide the amount of pure KHP by the amount of sample. 1.441 g / 1.864 g = 77.3% KHP.
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