Question

A pure 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.

Answer 1

0.0967 M;

0.128 g

Step-by-step 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