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

Answer:

$V_{base}=43.5cm^3$

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:

$V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}} \\\\V_{base}=\frac{45cm^3*0.29mol*dm^{-3}}{2*0.15mol*dm^{-3}} \\\\V_{base}=43.5cm^3$

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