Question

The Ksp of AgCl is 1.6 x 10-10. What is the solubility of AgCl in 0.0010 M FeCl3? Give your answer using scientific notation (1.23e-4) and to 2 significant figures (i.e., one decimal place). [a]

Answer 1

Answer:

$5.3\times 10^{-8}M$ is the solubility of AgCl in 0.0010 M ferric chloride.

Explanation:

$FeCl_3\rightarrow Fe^{3+}+3Cl^-$

Concentration of ferric chloride = $[FeCl_3]=0.0010 M$

$[Cl^-]=3\times [FeCl_3]=3\times 0.0010 M=0.0030 M$

Solubility product of silver chloride :$K_{sp}=1.6\times 10^{-10}$

$AgCl\rightarrow Ag^++Cl^-$

              S        (S+0.0030 )

The expression of a solubility product is given as:

$K_{sp}=[Ag^+][Cl^-]$

$1.6\times 10^{-10}=[S][S+0.0030]$

Solving fro S:

$S=5.3\times 10^{-8}M$

$5.3\times 10^{-8}M$ is the solubility of AgCl in 0.0010 M ferric chloride.

Answer 2

The solubility of AgCl in 0.0010 M FeCl3 is mathematically given as

S=5.3* 10^{-8}M

What is the solubility of AgCl in 0.0010 M FeCl3?

Question Parameter(s):

The Ksp of AgCl is 1.6 x 10-10

The solubility of AgCl in 0.0010 M FeCl3

Generally, the equation for the Chemical Reaction   is mathematically given as

FeCl3 ----> Fe^{3+}+3Cl^-

Therefore, Conc of  FeCl3

[Cl^-]=3* [FeCl_3]

[Cl^-]=3* 0.0010 M

[Cl^-]=0.0030 M

In conclusion, Ksp

Ksp=[Ag^+][Cl^-]

1.6* 10^{-10}=[S][S+0.0030]

S=5.3* 10^{-8}M

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