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JEE Exam  >  JEE Questions  >  Equal volumes of 0.02 M AgNO3and 0.02 M HCN w... Start Learning for Free
Equal volumes of 0.02 M AgNO3 and 0.02 M HCN were mixed. Calculate [Ag+] at equilibrium. Take Ka(HCN) = 9 × 10-10, Ksp (AgCN) =4 × 10-6.
  • a)
    [Ag+] = 5.699 × 10-5 M 
  • b)
     [Ag+] = 6.669 × 10-5 M 
  • c)
    [Ag+] = 11.66 × 10-5 M 
  • d)
    [Ag+] = 12.669 × 10-5 M
Correct answer is option 'B'. Can you explain this answer?
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Equal volumes of 0.02 M AgNO3and 0.02 M HCN were mixed. Calculate [Ag+...
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Equal volumes of 0.02 M AgNO3and 0.02 M HCN were mixed. Calculate [Ag+...
To solve this problem, we need to use the concept of equilibrium and the equilibrium constant. Let's break down the steps to find the concentration of Ag in the solution at equilibrium.

Step 1: Write the balanced chemical equation for the reaction between AgNO3 and HCN.
AgNO3(aq) + HCN(aq) ⇌ AgCN(s) + HNO3(aq)

Step 2: Write the expression for the equilibrium constant, K, for the reaction.
K = [Ag+][CN-]/[HCN]

Step 3: Determine the initial concentrations of Ag+ and HCN.
Since equal volumes of 0.02 M AgNO3 and 0.02 M HCN are mixed, the initial concentrations of Ag+ and HCN are both 0.02 M.

Step 4: Set up an ICE (Initial, Change, Equilibrium) table to track the changes in concentrations.
Initial:
[Ag+] = 0.02 M
[CN-] = 0 M
[HCN] = 0.02 M

Change:
[Ag+] = -x
[CN-] = +x
[HCN] = -x

Equilibrium:
[Ag+] = 0.02 - x M
[CN-] = x M
[HCN] = 0.02 - x M

Step 5: Substitute the equilibrium concentrations into the equilibrium constant expression.
K = [Ag+][CN-]/[HCN]
K = (0.02 - x)(x)/(0.02 - x)

Step 6: Solve for x using the given equilibrium constants.
K = 4 x 10^-6 = (0.02 - x)(x)/(0.02 - x)

Since the value of Ka(HCN) = 9 x 10^-10 is much smaller than Ksp(AgCN) = 4 x 10^-6, we can assume that x is very small compared to 0.02.

Simplifying the equation:
4 x 10^-6 = (0.02)(x)/(0.02)
4 x 10^-6 = x

Step 7: Calculate the concentration of Ag+ at equilibrium.
[Ag+] = 0.02 - x
[Ag+] = 0.02 - 4 x 10^-6
[Ag+] = 0.02 - 0.000004
[Ag+] = 0.019996 M

Step 8: Convert the concentration to scientific notation.
[Ag+] = 1.9996 x 10^-2 M

Therefore, the correct answer is option 'B': [Ag+] = 6.669 x 10^-5 M.
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Equal volumes of 0.02 M AgNO3and 0.02 M HCN were mixed. Calculate [Ag+] at equilibrium. Take Ka(HCN) = 9 × 10-10, Ksp(AgCN) =4 × 10-6.a)[Ag+] = 5.699 × 10-5Mb)[Ag+] = 6.669 × 10-5Mc)[Ag+] = 11.66 × 10-5Md)[Ag+] = 12.669 × 10-5MCorrect answer is option 'B'. Can you explain this answer?
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Equal volumes of 0.02 M AgNO3and 0.02 M HCN were mixed. Calculate [Ag+] at equilibrium. Take Ka(HCN) = 9 × 10-10, Ksp(AgCN) =4 × 10-6.a)[Ag+] = 5.699 × 10-5Mb)[Ag+] = 6.669 × 10-5Mc)[Ag+] = 11.66 × 10-5Md)[Ag+] = 12.669 × 10-5MCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Equal volumes of 0.02 M AgNO3and 0.02 M HCN were mixed. Calculate [Ag+] at equilibrium. Take Ka(HCN) = 9 × 10-10, Ksp(AgCN) =4 × 10-6.a)[Ag+] = 5.699 × 10-5Mb)[Ag+] = 6.669 × 10-5Mc)[Ag+] = 11.66 × 10-5Md)[Ag+] = 12.669 × 10-5MCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Equal volumes of 0.02 M AgNO3and 0.02 M HCN were mixed. Calculate [Ag+] at equilibrium. Take Ka(HCN) = 9 × 10-10, Ksp(AgCN) =4 × 10-6.a)[Ag+] = 5.699 × 10-5Mb)[Ag+] = 6.669 × 10-5Mc)[Ag+] = 11.66 × 10-5Md)[Ag+] = 12.669 × 10-5MCorrect answer is option 'B'. Can you explain this answer?.
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