Answer to Question #143858 in Chemistry for Sunny Bae

Question

A 0.1036-g sample containing only BaCl2 and NaCl is dissolved in 50 mL of
distilled water. Titrating with 0.07916 M AgNO3 requires 19.46 mL to reach the
Fajans endpoint. Report the %w/w BaCl2 in the sample.

Answer 1

During the titration, silver ions react with chloride anions and form silver chloride:

Ag⁺ + Cl^- $\rightarrow$ AgCl$\downarrow$ .

As it can be seen from this equation, at the equivalence point, the number of the moles of silver ions added to the sample and the number of the moles of chloride anions initially present in the sample are equal:

$n(Ag^+) = n(Cl^-)$ .

The number of the moles of silver cations added is:

$n(Ag^+) = cV = 0.07916\text{M}·19.46·10^{-3}\text{L}$

$n(Ag^+) = 1.5405·10^{-3}$ mol.

Therefore, 1.5405·10^-3 mol of chloride anions is contained in the sample.

The number of the moles of Cl^- in the sample is the sum of the number of the moles of NaCl and twice the number of the moles of BaCl² (as 1 molecule of BaCl² contains 2 chlorine anions):

$n(Cl^-) = n(NaCl) + 2n(BaCl_2)$ .

The number of the moles of NaCl and BaCl² can be calculated as the mass divided by the molar mass. The molar mass of NaCl is 58.44 g/mol and that of BaCl₂ is 208.23 g/mol.

If $w$ is the %w/w of BaCl₂ in the sample, then the number of the moles of Cl^- in the sample is:

$n(Cl^-) =2 \frac{w}{100\%}\frac{m(sample)}{M(BaCl_2)} + (1-\frac{w}{100\%})\frac{m(sample)}{M(NaCl)}$

$1.5405·10^{-3} =2 \frac{w}{100\%}\frac{0.1036}{208.23} + (1-\frac{w}{100\%})\frac{0.1036}{58.44}$

$\frac{w}{100} = \frac{1.5405·10^{-3} - \frac{0.1036}{58.44}}{2\frac{0.1036}{208.23} - \frac{0.1036}{58.44} } = 0.299$

Finally, the %w/w of BaCl₂ in the sample is 29.9%.

Answer: the %w/w of BaCl₂ in the sample is 29.9 %