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One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample with a silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate.

Suppose an EPA chemist tests a 250 mL sample of groundwater known to be contaminated with tin(II) chloride, which reacts with silver nitrate solution as follows:

[tex]\[ SnCl_2(aq) + 2AgNO_3(aq) \rightarrow 2AgCl(s) + Sn(NO_3)_2(aq) \][/tex]

The chemist adds a [tex]\(66.0 \, mM\)[/tex] silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate, finding that she has collected 3.8 mg of silver chloride.

Calculate the concentration of tin(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits.

Sagot :

GinnyAnswer
To calculate the concentration of tin(II) chloride contaminant in the original groundwater sample, we can follow these steps:

1. Convert the mass of silver chloride from mg to grams:
[tex]\[ 3.8 \text{ mg} = \frac{3.8}{1000} \text{ g} = 0.0038 \text{ g} \][/tex]

2. Convert the volume of the groundwater sample from mL to L:
[tex]\[ 250 \text{ mL} = \frac{250}{1000} \text{ L} = 0.25 \text{ L} \][/tex]

3. Calculate the molar mass of silver chloride (AgCl):
- Molar mass of Ag: [tex]\(107.8682 \, \text{g/mol}\)[/tex]
- Molar mass of Cl: [tex]\(35.45 \, \text{g/mol}\)[/tex]
[tex]\[ \text{Molar mass of AgCl} = 107.8682 \, \text{g/mol} + 35.45 \, \text{g/mol} = 143.3182 \, \text{g/mol} \][/tex]

4. Calculate the moles of silver chloride (AgCl):
[tex]\[ \text{Moles of AgCl} = \frac{0.0038 \, \text{g}}{143.3182 \, \text{g/mol}} \approx 2.6514427337211885 \times 10^{-5} \, \text{moles} \][/tex]

5. Determine the moles of tin(II) chloride (SnCl[tex]\(_2\)[/tex]) using the balanced chemical equation:
- From the balanced equation, 1 mole of SnCl[tex]\(_2\)[/tex] reacts with 2 moles of AgNO[tex]\(_3\)[/tex], producing 2 moles of AgCl.
- Therefore, the moles of SnCl[tex]\(_2\)[/tex] is half the moles of AgCl:
[tex]\[ \text{Moles of SnCl}_2 = \frac{2.6514427337211885 \times 10^{-5}}{2} \approx 1.3257213668605943 \times 10^{-5} \, \text{moles} \][/tex]

6. Calculate the concentration of tin(II) chloride (SnCl[tex]\(_2\)[/tex]) in the groundwater sample:
[tex]\[ \text{Concentration of SnCl}_2 = \frac{\text{Moles of SnCl}_2}{\text{Volume of groundwater sample in L}} = \frac{1.3257213668605943 \times 10^{-5}}{0.25} \approx 5.302885467442377 \times 10^{-5} \, \text{M} \][/tex]

7. Convert this concentration to millimolar (mM):
[tex]\[ \text{Concentration of SnCl}_2 \text{ in mM} = 5.302885467442377 \times 10^{-5} \times 1000 \approx 0.053028854674423774 \, \text{mM} \][/tex]

8. Round the answer to 2 significant digits:
[tex]\[ \text{Concentration of SnCl}_2 = 0.053 \, \text{mM} \][/tex]

Therefore, the concentration of tin(II) chloride contaminant in the original groundwater sample is approximately [tex]\(0.053\)[/tex] mM.
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