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Physical Chemistry

when 15ml of 0.5M AgNO3 is mixed with 45mL of 0.03M K2CrO4 . Find whether the precipitation occurs or not. Ksp pf Ag2CrO4=1.9×10^-12

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8 Years agoGrade
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ApprovedApproved Tutor Answer1 Year ago

To determine whether precipitation occurs when mixing 15 mL of 0.5 M AgNO3 with 45 mL of 0.03 M K2CrO4, we need to calculate the concentrations of the ions in the resulting solution and compare the ionic product (Q) to the solubility product constant (Ksp) of Ag2CrO4. If Q exceeds Ksp, precipitation will occur.

Step 1: Calculate the moles of each reactant

First, we find the number of moles of AgNO3 and K2CrO4 in the respective volumes:

  • Moles of AgNO3 = Concentration × Volume = 0.5 M × 0.015 L = 0.0075 moles
  • Moles of K2CrO4 = Concentration × Volume = 0.03 M × 0.045 L = 0.00135 moles

Step 2: Determine the total volume of the mixture

The total volume after mixing is:

  • Total Volume = 15 mL + 45 mL = 60 mL = 0.060 L

Step 3: Calculate the concentrations of Ag+ and CrO42-

Next, we calculate the concentrations of the ions in the final solution:

  • Concentration of Ag+ = Moles of AgNO3 / Total Volume = 0.0075 moles / 0.060 L = 0.125 M
  • Concentration of CrO42- = Moles of K2CrO4 / Total Volume = 0.00135 moles / 0.060 L = 0.0225 M

Step 4: Calculate the ionic product (Q)

The precipitation of silver chromate (Ag2CrO4) can be represented by the following equilibrium:

2 Ag+ + CrO42- ⇌ Ag2CrO4 (s)

The expression for the ionic product (Q) is given by:

Q = [Ag+]2 × [CrO42-]

Substituting the concentrations we found:

  • Q = (0.125)2 × (0.0225) = 0.015625 × 0.0225 = 0.0003515625

Step 5: Compare Q to Ksp

Now we compare Q to the Ksp of Ag2CrO4, which is given as 1.9 × 10-12:

  • Q = 0.0003515625
  • Ksp = 1.9 × 10-12

Since Q (0.0003515625) is much greater than Ksp (1.9 × 10-12), this indicates that the solution is supersaturated with respect to Ag2CrO4.

Final Thoughts

Given that Q exceeds Ksp, precipitation of silver chromate (Ag2CrO4) will indeed occur when these two solutions are mixed. This is a classic example of how solubility product constants can be used to predict the formation of precipitates in chemical reactions.