UmL36. For the reaction between 200.0 mL of 0.100 M silver nitrate and 95.00 mL of 0.100 Msodium chloride, the concentration of the silver ions in solution after the reaction is complete is:a. 0.00 M. All of the silver ions are used to make the precipitate.b. 0.100 M. The silver ions do not participate in the chemical reaction and are consideredspectator ions.c. greater than 0.100 M. For every 1.0 mol of silver nitrate there are 2.0 mol of silver ions inthe solution.d. less than 0.100 M. Even though silver ions do not react, they are diluted when mixed withthe sodium chloride solution.e. less than 0.100 M. Some, but not all, of the silver ions are used to make the precipitate.

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Asked Dec 2, 2019
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UmL
36. For the reaction between 200.0 mL of 0.100 M silver nitrate and 95.00 mL of 0.100 M
sodium chloride, the concentration of the silver ions in solution after the reaction is complete is:
a. 0.00 M. All of the silver ions are used to make the precipitate.
b. 0.100 M. The silver ions do not participate in the chemical reaction and are considered
spectator ions.
c. greater than 0.100 M. For every 1.0 mol of silver nitrate there are 2.0 mol of silver ions in
the solution.
d. less than 0.100 M. Even though silver ions do not react, they are diluted when mixed with
the sodium chloride solution.
e. less than 0.100 M. Some, but not all, of the silver ions are used to make the precipitate.
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UmL 36. For the reaction between 200.0 mL of 0.100 M silver nitrate and 95.00 mL of 0.100 M sodium chloride, the concentration of the silver ions in solution after the reaction is complete is: a. 0.00 M. All of the silver ions are used to make the precipitate. b. 0.100 M. The silver ions do not participate in the chemical reaction and are considered spectator ions. c. greater than 0.100 M. For every 1.0 mol of silver nitrate there are 2.0 mol of silver ions in the solution. d. less than 0.100 M. Even though silver ions do not react, they are diluted when mixed with the sodium chloride solution. e. less than 0.100 M. Some, but not all, of the silver ions are used to make the precipitate.

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Expert Answer

Step 1

Given,

Volume of AgNO3 = 200.0 mL = 0.2 L   (1mL = 0.001 L)

Molarity of AgNO3 = 0.100 M = 0.1 mol/L

Volume of NaCl = 95.00 mL = 0.095 L  

Molarity of NaCl = 0.100 M = 0.1 mol/L

No. of moles of AgNO3 and NaCl can be calculated as:

Moles of AgNO3= Molarity x Volume in litres
Moles of AgNO3 0.1 mol/L x 0.2 L 0.02 mol
Moles of NaCl = Molarity x Volume in litres
Moles of Nacl = 0.1 mol/L x 0.095 L = 0.0095 mol
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Moles of AgNO3= Molarity x Volume in litres Moles of AgNO3 0.1 mol/L x 0.2 L 0.02 mol Moles of NaCl = Molarity x Volume in litres Moles of Nacl = 0.1 mol/L x 0.095 L = 0.0095 mol

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Step 2

The balanced chemical reaction of silver nitrate and sodium chloride can be written as :

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

Thus, a precipitate is formed AgCl (s). The Ag+ ions that will be remaining at the end of this reactions will be the coming only from AgNO3 (aq).

AgNO3 (aq) → Ag+ (aq) + NO3- (aq)

Step 3

From the reaction, it is evident that, 1 mole of AgNO3 reacts with 1 mole of NaCl. Thus, the molar ratio of AgNO3 and N...

Moles of AgNO3 reacted 0.0095 mol
Moles of AgNO3 after the reaction is completed 0.02 0.0095 mol -0.0105 mol
. Moles of Ag ions after the reaction is completed 0.0105 mol
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Moles of AgNO3 reacted 0.0095 mol Moles of AgNO3 after the reaction is completed 0.02 0.0095 mol -0.0105 mol . Moles of Ag ions after the reaction is completed 0.0105 mol

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