Question

An unknown sample (0.6855 g) containing KHP is titrated to the equivalence point with 9.7 mL of 0.0789 M NaOH.

a) How many moles of KHP are present?

b) How many grams of KHP are present (molecular weight of KHP = 204.23 g/mol)?

c) What is the percent KHP in the sample?

Answer 1

The balanced Equation is :
KHC8H4O4(aq) + NaOH(aq) --------> KNaC8H4O4(aq) + H2O(l)
That means 1 mole of KHP reacts with 1 mole of NaOH
No . of moles of NaOH reacted , n = Molarity * Volume in L
= 0.0789 M * 0.0097 L
= 0.000765 moles
1 mole of KHP reacts with 1 mole of NaOH
0.000765 moles of KHP reacts with 0.000765 moles of NaOH ---(a)
mass of KHP ,m = No . of moles * Molar mass
= 0.000765 moles * 204.23 g / mol
= 0.1563 g ----(b)
% KHP = ( mass of KHP / mass of sample ) *100
= ( 0.1563 g / 0.6855 g ) *100
= 22.8 % ----(c)

Answer 2

1 mol of KHP reacts with 1 mol of NaOH

so at equivalence point KHP moles = NaOH moles

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We know that mole = mass/molar mass

                   ? mass = moles*molar mass

                              = (0.000765 mol)(204.23 g/mol)

                              = 0.156 g

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     Percent KHP in the sample = (0.156 g/0.6855 g)*100

                                            = 22.8%