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Well, now let's look at the common ion effect. This says that salt solubility is reduced if one of the component ions is already in solution. If the Ksp of AgCl is 1.8 x 10 to the -10, and we add AgCl to an existing 0.1 molar NaCl solution, what is the molar solubility of the Ag+ ion?

What would the solubility of the silver ion have been if the AgCl was added to pure water instead? KSP = 1.8 times 10 to- 10. This also equals the concentration of the silver times the concentration of the chlorine. Chlorine is essentially 0.1 molar.

So our silver will be given to us if we just solve this equation. We divide out the 0.1, Ag = 1.8 times 10 to the -9 molar. Now if we were going to be rigorous about this solution, we would then say, since we have 1.8 x 10 to the -9 molar silver from the AgCl. And remember that AgCl goes to Ag+ + Cl-, we would also have, 1.8 times 10 to -9 molar chlorines added to 0.1 molar Cl from the previous solution.

And then we would plug this value back into here and resolve for the silver. However, 0.1 is much larger than 1.8 times 10 to the -9. So we don't have to do this iteration. If the numbers were close, we would. Now what if we had put the silver chloride into pure water? Well, we said Ksp = 1.8 times 10 to the -10 = [Ag+][Cl-].

There's no existing chloride in the pure water, so we can just use essentially x times x, and say that Ksp which equals 1.8 times 10 to the -10 equals x squared. x, which equals a silver concentration equals 1.34 times 10 to the -5. How much more silver dissolves into the pure water than in the sodium chloride solution?

Let's do a simple division here. 1.34 times 10 to the -5 / 1.8 times 10 to -9. Now we can ignore this for the time being just to get a check on magnitude. And just assume it was 10 to the -5 / 10 to the -9 = 10 to the (-5- -9) or 10 to the 4th power, which is essentially a 1 with four 0s behind it.

There's around 10,000 times more silver ions in the pure water than there is in the sodium chloride solution. If we were to do an exact, we would get 7,453 times as much, but again, often the answers are in orders of magnitude. So as a quick check, just to make you're in the ballpark, you would check the order of magnitude.

And if you could find the right answer that way, you don't even bother doing the more complicated division.

Show TranscriptWhat would the solubility of the silver ion have been if the AgCl was added to pure water instead? KSP = 1.8 times 10 to- 10. This also equals the concentration of the silver times the concentration of the chlorine. Chlorine is essentially 0.1 molar.

So our silver will be given to us if we just solve this equation. We divide out the 0.1, Ag = 1.8 times 10 to the -9 molar. Now if we were going to be rigorous about this solution, we would then say, since we have 1.8 x 10 to the -9 molar silver from the AgCl. And remember that AgCl goes to Ag+ + Cl-, we would also have, 1.8 times 10 to -9 molar chlorines added to 0.1 molar Cl from the previous solution.

And then we would plug this value back into here and resolve for the silver. However, 0.1 is much larger than 1.8 times 10 to the -9. So we don't have to do this iteration. If the numbers were close, we would. Now what if we had put the silver chloride into pure water? Well, we said Ksp = 1.8 times 10 to the -10 = [Ag+][Cl-].

There's no existing chloride in the pure water, so we can just use essentially x times x, and say that Ksp which equals 1.8 times 10 to the -10 equals x squared. x, which equals a silver concentration equals 1.34 times 10 to the -5. How much more silver dissolves into the pure water than in the sodium chloride solution?

Let's do a simple division here. 1.34 times 10 to the -5 / 1.8 times 10 to -9. Now we can ignore this for the time being just to get a check on magnitude. And just assume it was 10 to the -5 / 10 to the -9 = 10 to the (-5- -9) or 10 to the 4th power, which is essentially a 1 with four 0s behind it.

There's around 10,000 times more silver ions in the pure water than there is in the sodium chloride solution. If we were to do an exact, we would get 7,453 times as much, but again, often the answers are in orders of magnitude. So as a quick check, just to make you're in the ballpark, you would check the order of magnitude.

And if you could find the right answer that way, you don't even bother doing the more complicated division.