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Question:A solution of AgNO3 (45 mL/0.45 M) was mixed with solution of NaCl (85 mL/1.35 x 10-2 M)
a) Calculate the ion product of the potential precipitate.

My answer:

mol AgNO3 = M x L
= 0.45 M x 0.045 M
= 0.02025

mol NaCl = M x L
= 0.0135 M x 0.085 L
= 0.0011475

(Ag^+) = mols / total volume
= 0.02025 / 0.130 L
= 0.1558

total volume = 85 mL + 45 mL
= 130 mL
= 0.130 L

(Cl^-) = mols / total volume
= 0.0011475 / 0.130 L
= 0.00883

Ion product = [Ag^+][Cl^-]
= [0.1558][0.00883]
= 0.001375

Is my answer correct?

  • chemistry -

    It looks ok to me. One point I want to make here. Technically, the Ksp of a salt, in this case, AgCl, cannot be exceeded (unless a supersaturated solution is formed). That is, (Ag^+)(Cl^-) can not be larger than Ksp which is about 1.8 x 10^-10. That's why what you have done is called the ion product. Since the ion product exceeds Ksp, you know a ppt will occur. How much AgCl will ppt. Enough AgCl will come out of solution so that the final (Ag^+) x the final (Cl^-) will be equal to 1.8 x 10^-10. In a saturated solution, (Ag^+) = (Cl^-) = 1.34 x 10^-5 molar so most will ppt. The point I want to make is that the ion product thing is an artificial one in that Ksp can't be exceeded. But the ion product is a way of comparing Ksp and knowing if a ppt will occur.

  • chemistry -

    Thanks for the tip

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