A volume of 68.375 ml of .0120M HNO₃ is required to completely neutralize 54.7 ml of 0.150M Ba(OH)₂.
First let us write a balanced chemical equation of the reaction of HNO₃ and Ba(OH)₂,
HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + H₂O
As we can see from the reaction,
One mole of HNO₃ reacts with one mole of Ba(OH)₂ to form the product.
It is given that 54.7ml of 0.150M Ba(OH)₂ with 0.120M sample of HNO₃,
So, we can write,
One mole HNO₃ = One mole Ba(OH)₂
We know,
Molarity = moles/volume of solution
So,
Moles = Molarity x volume
Hence,
Putting the values for Ba(OH)₂ and HNO₃,
0.120 x V = 0.150 x 54.7
V is the volume of HNO₃ required,
V = 68.375 ml.
The volume of HNO₃ required is 68.375 ml.
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