A volume of 100 mL of a 0.590 M HNO3 solution is titrated with 0.660 M...

Question

Question

A volume of 100 mL of a 0.590 M HNO3 solution is titrated with 0.660 M KOH. Calculate the volume of KOH required to reach the equivalence point.

Answer 1

Answer #1

Answer

89.39ml

Explanation

HNO3(aq) + KOH(aq) ---------> KNO3(aq) + H2O(l)

Stoichiometrically, 1mole of HNO3 reacts with 1mole of KOH

molarity = numbet of moles of solute per liter of solution

given moles of HNO3 = (0.590mol/1000ml) × 100ml = 0.0590mol

0.0590moles of HNO3 react with 0.0590moles of KOH

Volume of KOH solution containing 0.0590moles of KOH = (1000ml/0.660mol) × 0.0590mol= 89.39ml

Answer 2

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