Hello Guys!!
I got this question in class:
Given that the solubility product value (Ksp) of AgCl is 1.8 X 10[-10], will a precipitate form when 1cm[3] of 0.001M AgNO[3] is added to 1dm[3] of 0.001M NaCl?
I say NO. From the solubilty product value of AgCl, the solubility of AgCl is:
1.8 X 10[-10] = [Ag
- X [ Cl[-] i therefore found the square root of 1.8 X 10[-10] which gives the solubility of AgCl as 1.34 X 10[-5]moles per dm[3]
Now, how much Ag
- do we have in 1 cm[3] of 0.001M AgNO[3]?
since 1000 cm[3] AgNO[3] contains 0.001 moles
then 1 cm[3] AgNO[3] will contain 1 / 1000 X 0.001
= 1.0 X 10[-6] moles [Ag
if this amount of [Ag
- is place in the 1 dm[3] of NaCl - assuming there is no change in volume - then the new concentration of [Ag
- is 1.0 X 10[-6] moles per dm[3].
I therefore say: since the new concentration of Ag
- is less than the solubility of AgCl - then [NO PRECIPITATE WILL BE FORMED].
Am I right? My teacher and classmates don't think so.