Question

% composition = (m KHP/m unknown) X 100%. You've weighed out 1.6392 g of an impure...

% composition = (m KHP/m unknown) X 100%.

You've weighed out 1.6392 g of an impure KHP sample. The sample was dissolved in 50mL of DI water and titrated with 0.1139 M NaOH. If 29.37 mL of NaOH was needed to reach the endpoint, then what is the percent composition of KHP in your unknown sample?

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Answer #1

Balanced chemical reaction between KHP and NaOH

NaOH + KHC8H4O4 KNaC8H4O4 + H2O

no. of mole = molarity X volume of solution in liter

no. of mole of NaOH = 0.1139 X 0.02937 = 0.003345 mole

According to balanced chemical reaction 1 mole of NaOH react with 1 mole of KHP then to react with 0.003345 mole of NaOH required KHP = 0.003345 mole thus KHP in sample = 0.003345 mole

molar mass of KHP = 204.22 g/mol then 0.003345 mole of KHP = 0.003345 X 204.22 = 0.683 gm

in 1.6392 gm of impure KHP sample present 0.683 gm KHP

1.6392 gm impure KHP = 100 % sample then 0.683 gm KHP = 0.683 X 100 / 1.6392 = 35.23 %

percent composition of KHP in unknown sample = 35.23 %

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