A volume of 70.0 mL of a 0.820 M HNO3 solution is titrated with 0.900 M...

A volume of 70.0 mL of a 0.820 M HNO3 solution is titrated with 0.900 M KOH. Calculate the volume of KOH required to reach the equivalence point.

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Expert Solution

Consider reaction, HNO ₃ (aq) + KOH (aq) KNO ₃ (aq) + H₂O (l)

From reaction , 1 mol HNO ₃ 1 mol KOH 1 mol KNO ₃

We have relation, M _acid V _acid = M _base V _base

M _{HNO 3} V _{HNO 3} = M _KOH V _KOH

V _KOH = M _{HNO 3} V _{HNO 3} / M _KOH

V _KOH = 0.820 M 70.0 ml / 0.900 M

V _KOH = 63.77 ml

ANSWER : Volume of KOH required to reach the equivalence point = 63.8 ml

queen_honey_blossom answered 1 year ago

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