Apure sample of barium hydroxide of mass 8.37 g was dissolved and diluted to the mark in a 250 ml volumetric flask. it was found that 5.18 ml of this solution was needed to reach the stoichiometric point in a titration of 21 ml of a nitric acid solution. calculate the molarity of the hno3 solution. answer in units of m. 007 (part 2 of 2) 10.0 points what mass of hno3 was in the sample that was titrated with aqueous barium hydroxide? answer in units of g.

0.0967 M;

0.128 g

Explanation:

Barium hydroxide has formula Ba(OH)₂, and molar mass equal to 171.34 g/mol, so in 8.37 g there is:

n = 8.37/171.34 = 0.049 mol

The concentration in the volumetric flask will be:

M = 0.049/0.250 = 0.196 M

So, in 5.18 mL (0.00518 L) of the solution, there is:

n = 0.196*0.00518 = 1.015x10⁻³ mol which reacts tottaly with HNO₃ in the reaction:

Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O

So, for the stoichiometry:

1 mol of Ba(OH)₂   2 moles of HNO₃

1.015x10⁻³ mol of Ba(OH)₂   x

By a simple direct three rule:

x = 2.030x10⁻³ mol of HNO₃

The molarity is the number of moles divided by the volume (21 mL = 0.021 L)

M(HNO₃) = 2.030x10⁻³/0.021 = 0.0967 M

The molar mass of HNO₃ is 63.01 g/mol, so the mass (m) is the molar mass multiplied by the number of moles:

m = 63.01*2.030x10⁻³

m = 0.128 g

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