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If 20.00 mL of a 0.0090 M solution of (NH4)2S is mixed with 120.00 mL of a
0.0082 M solution of Al(NO3)3, does a precipitate form? The Ksp of Al2S3 is
2.00*10^-7. Included calculated ion product in answer.

Sagot :

Sebassandinidn

Answer:

No, no precipitate is formed.

Explanation:

Hello there!

In this case, since the reaction between ammonium sulfide and aluminum nitrate is:

[tex]3(NH_4)_2S(aq)+2Al(NO_3)_3(aq)\rightarrow Al_2S_3(s)+6NH_4NO_3(aq)[/tex]

In such a way, we can calculate the concentration of aluminum and sulfide ions in the solution as shown below, and considering that the final total volume is 140.00 mL:

[tex][Al^3^+]=\frac{120.00mL*0.0082M}{140.00mL}=0.00703M[/tex]

[tex][S^2^-]=\frac{20.00mL*0.0090M}{140.00mL}=0.00129M[/tex]

In such a way, we can calculate the precipitation quotient by:

[tex]Q=[Al^3^+]^2[S^2^-]^3=(0.00703)^2(0.00129)^3=1.05x10^{-13}[/tex]

Which is smaller than Ksp and meaning that the precipitation does not occur.

Regards!

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