Question #50369

Calculate the volume of 0.55M HNO3 necessary to neutralize 5500 cm3 of 0.45M KOH

Expert's answer

For this reaction we can use an equation: M1V1 =M2V2.

So, we get: 0.55*V1 = 0.45*5500; V1 =0.45*5500/0.55 = 4500cm3 of HNO3.

So, we get: 0.55*V1 = 0.45*5500; V1 =0.45*5500/0.55 = 4500cm3 of HNO3.

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