If the endpoint in a titration was reached using a volume of 37.60 mL and the...

If the endpoint in a titration was reached using a volume of 37.60 mL and the concentration of the standard NaOH solution was 0.3214 M, how many grams of KHP were neutralized in this experiment? (MM of KHP = 204.23 g/mol)

Report your answer in grams but do not include units in your response.

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Expert Solution

Consider reaction, KHP + NaOH KNaP + H2O

According to reaction, 1 mol KHP reacts with 1 mol NaOH. Therefore, at equivalence point no. of moles of KHP = No. of moles of NaOH.

We have relation , [ NaOH ] = No. of moles of NaOH / volume of solution in L

No. of moles of NaOH = [ NaOH ] volume of solution in L

No. of moles of NaOH =0.3214 mol / L 0.03760 L

No. of moles of NaOH = 0.012085 mol

We have relation, No. of moles of NaOH = No. of moles of KHP.

No. of moles of KHP. = 0.012085 mol

We know that , No. of moles = Mass / Molar mass

Mass of KHP = No. of moles Molar mass

Mass of KHP = 0.012085 mol 204.23 g / mol ( Molar mass of KHP = 204.23 g /mol )

Mass of KHP = 2.468 g

ANSWER : 2.468 g KHP is neutralized in this experiment.

queen_honey_blossom answered 10 months ago

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