Question

Not a real experiment, just a homework question

In order to test the concentration of chlorine ions, Cl - (aq) in a water reservoir,

a sample of 300 ml was taken from a water reservoir to the laboratory

and a number of experiments were performed.

In the first stage, the sample was divided into two cups.

A 100 mL solution was prepared in the first glass, and in the second glass 200 ml

Several experiments were performed.

First cup - Add 100 ml of AgNO3 solution (aq) at a concentration of 0.8 M

Second cup - Add 200 mL NaCl solution (aq) at a concentration of 0.4M

A) In the first glass the following reaction occurred in its entirety:

Ag (aq) Cl- (aq) → AgCl (s)

1) How many mol of silver ions were inserted into the reaction?

2) How many g of sediment, AgCl (s) were received in response, in the first glass?

3) What is the chlorine ion concentration in the first glass before adding AgNO3?

B) after adding NaCl (aq) in the second glass

1) Was the concentration of chlorine ions greater? smaller?

Or equal to their concentration in the reservoir water?

2) What is the final concentration of chlorine ions in the solution after adding sodium chloride solution?

Answer 1

Part A:

(1) Ag⁺ + Cl^- $\rightarrow$ AgCl(s) the reaction is taking place to an entirety.

Moles of silver ions added = molarity of solution × volume (in L)

= 0.8mol/L × 100×10^-3L = 0.08mol (answer)

(2)

Moles of AgCl precipitated = moles of Ag⁺ taken = 0.08mol

Mass of AgCl precipitated = 0.08mol × 143.3213g/mol = 11.4657gm

Mass of sediment, AgCl received = 11.47g. (Answer)

(3)

Moles of Cl^- present in the sample = moles of silver ions added = 0.08mol

[Cl^-] = number of moles / volume of sample (in L) = 0.08mol/0.10L = 0.8mol/L = 0.8M. (Answer)

Part B:

(1) after addition of given NaCl, the concentration of chloride ion will be smaller than their concentration in reservoir .

As the added solution of NaCl has smaller concentration of chloride ion than original sample. Its effect will be similar to dilution.

(2)

Moles of Cl^- from the sample = 0.8mol/L × 0.200L = 0.16mol

Moles of Cl^- from NaCl = 0.4mol/L× 0.200L = 0.08mol

Total chloride moles = 0.16 + 0.08 = 0.24mol

Concentration of chloride ion = 0.24mol/0.400L = 0.6M (answer)