# One way the U. S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitratesolution.

One way the U. S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of 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 200. mL sample of groundwater known to be contaminated with nickel(II) chloride, which would react with
silver nitrate solution like this:
NiCl(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ni(NO3),(aq)
The chemist adds 73.0 mm silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the
precipitate. She finds she has collected 6.7 mg of silver chloride.
Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits.

## This Post Has 11 Comments

1. lilpetals says:

3.8x10⁻⁵ M

Explanation:

By the reaction given, the stoichiometry between FeCl3 and AgNO3 and AgCl is 1:3:3. The molar mass of AgCl is 143.32 g/mol, so the number of moles that was formed is:

n = mass/molar mass

n = 0.0036g/143.32g/mol

n = 2.51x10⁻⁵ mol

Thus, the number of moles of FeCl3 that had precipitated was:

1 mol of FeCl3 3 moles of AgCl

x 2.51x10⁻⁵ mol

By a simple direct three rule:

x = 8.37x10⁻⁶ mol

The precipitation is a reversible reaction, and, for the AgCl, is has an equilibrium constant Kps = 1.8x10⁻¹⁰, then, the solubility, S, (the maximum concentration of the ions that will not precipitate) can be calculated:

S² = Kps

S = √1.8x10⁻¹⁰

S = 1.34x10⁻⁵ mol/L

The volume of the sample was 250 mL = 0.250 L, so the number of moles of Cl⁻ that was not precipitated is:

n = 1.34x10⁻⁵ mol/L*0.250 L

n = 3.35x10⁻⁶ mol

Because of the stoichiometry:

1 mol of FeCl3 3 moles of Cl-

y 3.35x10⁻⁶ mol

By a simple direct three rule:

y = 1.12x10⁻⁶ mol of FeCl3

The total number of moles of FeCl3 is then:

n = x + y

n = 9.5x10⁻⁵ mol

The concentration is the number of moles dived by the volume:

9.5x10⁻⁵ /0.250 = 3.8x10⁻⁵ M

2. Expert says:

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

mark brainliest and have a great day!

3. Ellafrederick says:

1.1 × 10⁻⁴ M

Explanation:

Let's consider the following double displacement reaction.

CuCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s)+ Cu(NO₃)₂(aq)

We can establish the following relations:

The molar mass of AgCl is 143.32 g/mol.The molar ratio of AgCl to CuCl₂ is 2:1

The moles of CuCl₂ that reacted to produce 7.7 mg of AgCl are:

$7.7 \times 10^{-3} gAgCl.\frac{1molAgCl}{143.32gAgCl} .\frac{1molCuCl_{2}}{2molAgCl} =2.7 \times 10^{-5}molCuCl_{2}$

The molarity of CuCl₂ is:

$M=\frac{2.7 \times 10^{-5}molCuCl_{2}}{0.250L} =1.1\times 10^{-4} M$

4. mdbar321 says:

Explanation:

One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of 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 200.mL sample of groundwater known to be contaminated with copper(II) chloride, which would react with silver nitrate solution like this: CuCl2 (aq) + 2AgNO3 (aq) ? 2AgCl (s) + CuNO32 (aq) The chemist adds 26.0m M silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 6.1mg of silver chloride. Calculate the concentration of copper(II) chloride contaminant in the original groundwater sample. Be sure your answer has the correct number of significant digits.

5. Expert says:

i believe it is c.) 1575 j.

6. mathfan5049 says:

6.5 mg/L.

Explanation:

Step one: write out and Balance the chemical reaction in the Question above:

NiCl2 + 2AgNO3 > 2AgCl + Ni(NO3)2.

Step two: Calculate or determine the number of moles of AgCl.

So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:

Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.

Step three: Calculate or determine the number of moles of NiCl2.

Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.

Step four: detemine the mass of NiCl2.

Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.

Step five: finally, determine the concentration of NiCl2.

1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.

7. fheight01 says:

The balanced chemical reaction is expressed as follows:

CuCl2 (aq) + 2AgNO3 (aq) → 2AgCl (s) + CuNO32 (aq)

To determine the concentration of copper(II) chloride contaminant in the original groundwater sample, we use the final amount of silver chloride that was produced from the reaction and the relation of the substances from the chemical reaction. We calculate as follows:

mmol AgCl = 6.1 mg AgCl ( 1 mmol / 143.35 mg ) = 0.0426 mmol
mmol CuCl2 = 0.0426 mmol AgCl ( 1 mmol CuCl2 / 2 mmol AgCl ) = 0.0213 mmol CuCl2
concentration of CuCl2 in the original water sample = 0.0213 mmol CuCl2 / 200.0 mL = 1.0638 x 10^-4 mmol / mL or 1.0638 x 10^-4 mol/L

8. hdjsjfjruejchhehd says:

4.1 × 10⁻⁵ M

Explanation:

Let's consider the following balanced equation.

FeCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Fe(NO₃)₃(aq)

We can establish the following relations:

The molar mass of AgCl is 143.32 g/mol.The molar ratio of AgCl to FeCl₃ is 3:1.

When 3.5 mg of AgCl are collected, the moles of of FeCl₃ that reacted are:

$3.5 \times 10^{-3} gAgCl.\frac{1molAgCl}{143.32gAgCl} .\frac{1molFeCl_{3}}{3molAgCl} =8.1 \times 10^{-6} molFeCl_{3}$

The sample has a volume of 200 mL (0.200 L). The molar concentration of FeCl₃ is:

$\frac{8.1 \times 10^{-6} molFeCl_{3}}{0.200L} =4.1 \times 10^{-5}M$

9. gshreya2005 says:

5.0x10⁻⁵ M

Explanation:

It seems the question is incomplete, however this is the data that has been found in a web search:

" One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose a EPA chemist tests a 250 mL sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:

NiCl₂ + 2AgNO₃ →  2AgCl + Ni(NO₃)₂

The chemist adds 50 mM silver nitrate solution to the sample until silver chloride stops forming. She then washes, dries, and weighs the precipitate. She finds she has collected 3.6 mg of silver chloride. Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits. "

Keep in mind that while the process is the same, if the values in your question are different, then your answer will be different as well.

First we calculate the moles of nickel chloride found in the 250 mL sample:

3.6 mg AgCl ÷ 143.32 mg/mmol * $\frac{1mmolNiCl_{2}}{2mmolAgCl}$ = 0.0126 mmol NiCl₂

Now we divide the moles by the volume to calculate the molarity:

0.0126 mmol / 250 mL = 5.0x10⁻⁵M

10. kacr7292 says:

15.0g/L is the concentration of nickel(II) chloride contaminant in the original groundwater sample.

Explanation:

Based on the reaction:

NiCl2 (aq) + 2AgNO3 (aq) -> 2AgCl (s) + Ni(NO3)2 (aq)

Where 1 mole of NiCl₂ reacts producing 2 moles of AgCl.

To solve this problem, we need to convert mass of AgCl to moles to know the moles of NiCl₂ that reacts. With these moles and the volume of the sample (250mL = 0.250L), we can determine the molar concentration of the contaminant in the sample

Moles AgCl:

8.3g of AgCl were collected. In moles (Molar mass AgCl: 143.32g/mol):

8.3g AgCl * (1mol / 143.32g) = 0.05791 moles AgCl

Moles NiCl₂:

As 2 moles of AgCl are produced from 1 mole of NiCl₂. Moles of NiCl₂ are:

0.05791 moles AgCl * (1 mole NiCl₂ / 2 moles AgCl) = 0.02896 moles NiCl₂

Molar concentration:

0.02896 moles NiCl₂ / 0.250L =

0.1158M

In g/L (Molar mass NiCl₂: 129.6g/mol):

0.1158 mol / L * (129.6g / mol) =

15.0g/L is the concentration of nickel(II) chloride contaminant in the original groundwater sample.

11. webbjalia04 says:

1.1 × 10⁻⁴ M

Explanation:

Let's consider the following balanced equation.

CuCl₂(aq)+ 2 AgNO₃(aq) ⇄ 2 AgCl(s) + Cu(NO₃)₂(aq)

The molar mass of AgCl is 143.32 g/mol. The moles corresponding to 7.7 mg (7.7 × 10⁻³ g) are:

7.7 × 10⁻³ g × (1 mol / 143.32 g) = 5.4 × 10⁻⁵ mol

The molar ratio of AgCl to CuCl₂ is 2:1. The moles of CuCl₂ are:

1/2 × (5.4 × 10⁻⁵ mol) = 2.7 × 10⁻⁵ mol

The molarity of CuCl₂ is:

M = 2.7 × 10⁻⁵ mol / 0.250 L = 1.1 × 10⁻⁴ M