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Equal volumes of 0.1 M AgNO₃ and 0.2 M NaCl are mixed. The concentration of NO₃- ions in the mixture will be
- a)0.1 M
- b)0.05 M
- c)0.2 M
- d)0. 15 M
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Equal volumes of 0.1 M AgNO₃ and 0.2 M NaCl are mixed. The conce...
Calculation of concentration of NO- ions in the mixture
Given:
- Volume of 0.1 M AgNO3 solution = Volume of 0.2 M NaCl solution
- Concentration of AgNO3 solution = 0.1 M
- Concentration of NaCl solution = 0.2 M
Step 1: Write the balanced chemical equation for the reaction that takes place when AgNO3 and NaCl are mixed.
AgNO3 + NaCl → AgCl + NaNO3
Step 2: Write the ionic equation for the reaction.
Ag+ + NO3- + Na+ + Cl- → AgCl + Na+ + NO3-
Step 3: Cancel out the spectator ions (ions that appear on both sides of the equation) to get the net ionic equation.
Ag+ + Cl- → AgCl
Step 4: Calculate the moles of Ag+ ions present in the mixture.
Moles of Ag+ ions = Molarity of AgNO3 x Volume of AgNO3 solution
= 0.1 x V
Step 5: Since the Ag+ and Cl- ions react in a 1:1 ratio, the moles of Cl- ions in the mixture will be equal to the moles of Ag+ ions.
Moles of Cl- ions = Moles of Ag+ ions
= 0.1 x V
Step 6: Calculate the total volume of the mixture.
Total volume = Volume of AgNO3 solution + Volume of NaCl solution
= 2V
Step 7: Calculate the concentration of Cl- ions in the mixture.
Concentration of Cl- ions = Moles of Cl- ions / Total volume
= (0.1 x V) / (2V)
= 0.05 M
Step 8: Since the NO3- ions are also present in the AgNO3 solution and do not react with NaCl, the concentration of NO3- ions in the mixture will be equal to the concentration of NO3- ions in the AgNO3 solution, which is 0.1 M.
Therefore, the concentration of NO- ions in the mixture will be 0.05 M (Option B).
Given:
- Volume of 0.1 M AgNO3 solution = Volume of 0.2 M NaCl solution
- Concentration of AgNO3 solution = 0.1 M
- Concentration of NaCl solution = 0.2 M
Step 1: Write the balanced chemical equation for the reaction that takes place when AgNO3 and NaCl are mixed.
AgNO3 + NaCl → AgCl + NaNO3
Step 2: Write the ionic equation for the reaction.
Ag+ + NO3- + Na+ + Cl- → AgCl + Na+ + NO3-
Step 3: Cancel out the spectator ions (ions that appear on both sides of the equation) to get the net ionic equation.
Ag+ + Cl- → AgCl
Step 4: Calculate the moles of Ag+ ions present in the mixture.
Moles of Ag+ ions = Molarity of AgNO3 x Volume of AgNO3 solution
= 0.1 x V
Step 5: Since the Ag+ and Cl- ions react in a 1:1 ratio, the moles of Cl- ions in the mixture will be equal to the moles of Ag+ ions.
Moles of Cl- ions = Moles of Ag+ ions
= 0.1 x V
Step 6: Calculate the total volume of the mixture.
Total volume = Volume of AgNO3 solution + Volume of NaCl solution
= 2V
Step 7: Calculate the concentration of Cl- ions in the mixture.
Concentration of Cl- ions = Moles of Cl- ions / Total volume
= (0.1 x V) / (2V)
= 0.05 M
Step 8: Since the NO3- ions are also present in the AgNO3 solution and do not react with NaCl, the concentration of NO3- ions in the mixture will be equal to the concentration of NO3- ions in the AgNO3 solution, which is 0.1 M.
Therefore, the concentration of NO- ions in the mixture will be 0.05 M (Option B).
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Community Answer
Equal volumes of 0.1 M AgNO₃ and 0.2 M NaCl are mixed. The conce...
[NO3-] in 0.1M AgNO3 =0.1M
When 0.1M AgNO3 is mixed with 0.2M NaCl the volume of the solution becomes double and that of conc of NO3- will be half.
So,
[NO3-] in the mixture will be = 0.1/2
=0.05M
So, answer is option B.
When 0.1M AgNO3 is mixed with 0.2M NaCl the volume of the solution becomes double and that of conc of NO3- will be half.
So,
[NO3-] in the mixture will be = 0.1/2
=0.05M
So, answer is option B.
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Equal volumes of 0.1 M AgNO₃ and 0.2 M NaCl are mixed. The concentration of NO₃- ions in the mixture will bea)0.1 Mb)0.05 Mc)0.2 Md)0. 15 MCorrect answer is option 'B'. Can you explain this answer?
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