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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#339914 on Dec 2023
#339914 on Dec 2023