Question

Calculate the volume of a 0.15 mol dm-3 Ba(OH)2 solution required to completely neutralize 45 cm3 of a 0.29 mol dm-3 HNO3 solution. Note: Ba(OH)2 + 2HNO3 --> Ba(NO3)2 + 2H2O

Answer 1

`V_(base)=43.5cm^3`

Step-by-step explanation:

Hello there!

In this case, given the neutralization chemical reaction of barium sulfate and nitric acid, it is possible to evidence the 1:2 mole ratio between them; thus, at the equivalence point we have:

`2n_(base)=n_(acid)`

Which in terms of molarities and volumes is:

`2M_(base)V_(base)=M_(acid)V_(acid)`

In such a way, by solving for the volume of base, we proceed as follows:

`V_(base)=(M_(acid)V_(acid))/(2M_(base)) \\\\V_(base)=(45cm^3*0.29mol*dm^(-3))/(2*0.15mol*dm^(-3)) \\\\V_(base)=43.5cm^3`

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