Answer to Question #116703 in Chemistry for Aly

Question

A 10.00-mL aliquot of a commercial algaecide contg. An
organomercury cpd. was treated with conc. HNO3 and
evaporated just to dryness. The residue of mercuric nitrate
was dissolved in dilute HNO3 & diluted to 250.0 mL with
water. A 52.00-mL aliquot of this soln was treated with
20.00 mL of 0.04966 M EDTA & mixed thoroughly for 20
min. After adjusting the pH 10, the x’ss EDTA req’d 18.04
mL of 0.04711 M MgCl2 for titration to the EBT end pt.
Calc. conc. of mercury in the original sample in units of mg
Hg / mL (200.59 g/mol).

Answer 1

Let's solve this problem from the end. The number of the moles of excess EDTA is the product of MgCl₂ concentration and volume used:

$n(EDTA)_{exc} = cV = 0.04711\text{ M}·18.04·10^{-3}\text{ L} = 0.8499$ mmol.

The number of the moles of EDTA added to aliquot:

$n(EDTA)_{tot} = cV = 0.04966\text{ M}·20.00·10^{-3}\text{ L} = 0.9932$ mmol.

Therefore, the number of the moles of EDTA used for the complexation with mercury is:

$n(EDTA)_{Hg} = 0.9932 - 0.8499 = 0.1433$ mmol.

As one ion of EDTA reacts with one ion of mercury, the number of the moles of Hg ions is equal to the number of the moles of EDTA used for their complexation. As only 52 mL aliquot was taken from 250 mL sample, the number of he moles of mercury must be multiplied by the factor:

$n(Hg^{2+}) = 0.1433·\frac{250}{52} = 0.6891$ mmol.

Therefore, the concentration of mercury in the original sample is:

$c = \frac{n}{V} = \frac{0.6891·10^{-3}\text{ mol}}{10.00·10^{-3}\text{ L}} = 0.06891$ M.

And in mg/mL units:

$c·M = 0.06891·200.59 = 13.82$ mg/mL.

Answer: the concentration of mercury in the original sample is 13.82 mg/mL.