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Stimulus 2 A student sets up a titration for the standardization of NaOH using KHP (KHC3H404). The reaction proceeds as follo
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Answer #1

a) We have to find the total volume of NaOH needed to neutralise KHP.

Initial reading of burette NaOH = 2.15 mL

Final reading of burette NaOH = 20.83 mL

Total reading = Final reading - Initial reading

= 20.83 - 2.15

= 18.68 mL (In 4 significant figures)

b) mass of KHP used = 0.3461 g

Molar mass of KHP = 204.2 g/mol

Number of moles = Given mass / molar mass

Number of moles of KHP =  0.3461g / 204.2 g/mol

= 0.001695 moles

The reaction between NaOH and KHP is in 1:1 stoichiometry. Therefore 0.001695 moles of KHP is must reacted with same number of moles od NaOH.

Number of moles of NaOH = 0.001695 moles

Volume of NaOH solution = 18.68 mL = (18.68/1000)L

= 0.01868 L

Molarity = no of moles / Volume in L

Molarity of NaOH = 0.001695moles 0.01868L

= 0.0907 M

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