Question

A student weighs a sample of potassium hydrogen phthalate (KHP) to prepare a primary standard for a titration. She later discovers that the KHP was contaminated with sugar. To determine the amount of KHP in the mixture, she takes 5.942 g of the mixture and make a 100.0 mL solution. The student then titrates 10.00 mL of this solution with a 0.1491 M sodium hydroxide solution. She finds that 13.12 mL of the NaOH solution is needed to reach the end point. What is the percent by mass of KHP in the mixture? percent by mass =

Answer 1

First question is answered here. Kindly post the other two questions separately.

1. KHP has one acidic hydrogen. Therefore the milli moles of NaOH used in titration is equal to milli moles of KHP present.

milli moles of NaOH = molarity (in moles/L) x (Volume in ml)

                                 = 0.1491 M x (13.12 ml)

                                 = 1.956 = milli moles of KHP.

So, the molarity of KHP (as 10 ml solution was taken ) = 1.956 / 10 = 0.1956 M

milli moles present in 100 ml = (0.1956 M) x 100 ml = 19.56

Molar mass of KHP (Molecular formula C₈H₅KO₄ ) is 204 g/mol

so, mg of KHP present = (19.56 milli moles) x (204 g/mol ) =3991

      or KHP present = 3.991 g

The percent by mass present in the sample =[ (3.991 g) / (5.942 g) ] x 100 = 67.16 %              ... Answer

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