A solution contains 2.0x10^-4 M Ag and 1.5x10^-3M Pb. If NaI is added, will AgI (Ksp = 8.3x10^-17) or PbI (Ksp= 7.9x10^-9) precipitate first? Specify the concentration of I needed to begin precipitation.
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2007-03-21 09:50:07 · 4 answers · asked by __ 3 in Science & Mathematics ➔ Chemistry
The solubility product of silver iodide is 8.3x10^-17 and that of lead iodide is 7.9x10^-9, therfore for exceeding the value of the solubility product less concentration of iodide ion will be required for silver iodide in comparison to lead iodide as shown below:
Ag(2.0x10^-4)* X > 8.3x10^-17;
or X > 8.3x10^-17/2.0x10^-4;
> 2.06x10^-13
For precipitation of lead iodide :
Pb(1.5x10^-3)*X > 7.9x10^-9
or X > 7.9x10^-9/ 1.5x10^-3
Or X > 5.3x10^-6.
Therefore, only 2.06x10^-13 M iodide is required to precipitate silver iodide against 5.3x10^-6 M iodide ion needed for precipitation of lead iodide.
2007-03-28 01:04:47 · answer #1 · answered by sb 7 · 0⤊ 1⤋
yes
2007-03-21 09:52:42 · answer #2 · answered by Anonymous · 1⤊ 7⤋
JUST PUT ON DEODERENT
2007-03-21 09:57:27 · answer #3 · answered by Anonymous · 0⤊ 6⤋
thanks 4 2
2007-03-21 09:53:45 · answer #4 · answered by just tht kid over there 3 · 0⤊ 6⤋