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A volume of 90.0 mL of a 0.820 M HNO3 solution is titrated with 0.220 M...

A volume of 90.0 mL of a 0.820 M HNO3 solution is titrated with 0.220 M KOH. Calculate the volume of KOH required to reach the equivalence point.

Express your answer to three significant figures and include the appropriate units.

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Homework Answers

Answer #1

HNO3+ KOH --------------------> KNO3+H2O

M1= 0.820M                                                                                         

V1= 90ml

n1=1 mole(from equation)

M2= 0.220M

V2= ?ml

n2=1 mole(from equation)

ACCORDING TO THE EQUATION

M1V1/n1= M2V2/n2

0.820*90/1= 0.220*v2/1

V2= 0.820*90/0.220= 335.45ml

335.45 ml of o.220M KOH is needed

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