Given #K_"sp"*AgCl=1.8xx10^(-10)#, if #[Ag^+]=10^(-10)# and #[Cl^-]=10^(-4)#, will silver chloride precipitate?

Answer 1

#Ag^(+) + Cl^(-) rightleftharpoonsAgCl(s)darr#

#K_"sp"=1.8xx10^-10#

We have the #K_"sp"# value, and thus we know that if the ion product #[Ag^+][Cl^-]>K_"sp"#, silver chloride will precipitate until #[Ag^+][Cl^-]-=K_"sp"#.
For #A.# #[Ag^+]=10^-10*mol*L^-1#; #[Cl^-]=10^-4*mol*L^-1#.
The ion product #[Ag^+][Cl^-]=10^-10xx10^-4=10^-14#,
Since this ion-product is LESS than #K_"sp"# silver chloride will not precipitate. I think you are due to do the next problems.

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Answer 2

Silver chloride will not precipitate because the ion product ([Ag^+][Cl^-]) is much larger than the K_sp value, indicating that the solution is unsaturated with respect to AgCl.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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