Question #150711

A pure sample of barium hydroxide of mass 5.29 g was dissolved and diluted to the mark in a 250 mL volumetric flask. It was found that 13.4 mL of this solution was needed to reach the stoichiometric point in a titration of 23.4 mL of a nitric acid solution. Calculate the molarity of the HNO3 solution. Answer in units of M.

Expert's answer

Equation of reaction

Ba(OH)2 + 2HNO3 -------- Ba(NO3)2+2H2O

To find the conc of Ba(OH)2 in mol/dm^{3 }we need the concentration in g/dm^{3}

To find the conc in g/dm^{3}

If 250mL contains 5.29g of Ba(OH)2

Then 1000mL will contain=5.29×1000/250

=21.16g/dm^{3}

Conin mol/dm^{3} = conc in g/dm^{3}÷molar mass

Conc of Ba(OH)2 in mol/dm^{3}=21.16/171

=0.124mol/dm^{3}

To calculate the conc of HNO3,we use the formula;

CaVa/na=CbVb/no

Where Ca=conc of acid Cb=conc of base

Va=vol of acid Vb=vol of base

na=no of mole of acid nb=no of mole of base

Making Ca the subject of the formula we have;

Ca=CbVbna/Vanb

Ca=0.124×13.4×2/1×23.4

Ca=0.142

Therefore the molarity of HNO3=0.142M

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