# 0.10 mol of #AgNO_3# is dissolved in #"1.00 L"# of #"1.00 M "NH_3#. If #"0.010 mol"# of #NaCl# is added, will #"AgCl (s)"# precipitate?

##
Given:

#K_f=1.6xx10^7#
#K_(sp)=1.8xx10^-10#

Apparently the calculated #Q_(sp)# , #9.8xx10^-11# , is less than given #K_(sp)# . What does this mean for the matter of 'precipitates or not' ?

Given:

#K_f=1.6xx10^7# #K_(sp)=1.8xx10^-10#

Apparently the calculated

##### 1 Answer

No. The reaction wants to consume the

The **complex formation** reaction of

#"Ag"^(+)(aq) + 2"NH"_3(aq) rightleftharpoons "Ag"("NH"_3)_2^(+)(aq)#

#"I"" "0.100" "" "" "1.00" "" "" "" "0#

#"C"" "-x" "" "" "-2x" "" "" "+x#

#"E"" "0.100 - x" "1.00-2x" "" "x#

It just so happens that

So, all we care about here is

#K_f = 1.6 xx 10^7 = (["Ag"("NH"_3)_2^+])/(["Ag"^(+)]["NH"_3]^2)#

#= x/((0.100 - x)(1.00 - 2x)^2)#

Since **huge**, we make the approximation that **PRODUCT-FAVORED**. This is an important fundamental concept that you should note.

NOTE:In doing so, we must make sure we do NOT plug it in for#[Ag^(+)]# . (Why? Well, just try it and evaluate the right-hand side.)

Hence:

#1.6 xx 10^7# #"M"^(-2) ~~ ("0.100 M")/((0.100 - x)("1.00 M" - 2("0.100 M"))^2)#

#= "0.15625 M"^(-1)/(0.100 - x)#

So now, we can solve for

#["Ag"^(+)]_(eq) = 0.100 - x = 9.77 xx 10^(-9) "M"#

#=> x ~~ "0.100 M"# as expected... just not EXACTLY#"0.100 M"# .

And as expected, the silver is almost all gone (but not quite). This silver cation can then react with the incoming

Now, if a precipitate forms, it means the reaction given by

#"AgCl"(s) rightleftharpoons "Ag"^(+)(aq) + "Cl"^(-)(aq)#

was previously **PRODUCT-FAVORED**, i.e

We officially have

- Is it at EQUILIBRIUM? If
#Q_(sp) = K_(sp)# , then yes... - Is it PRODUCT-FAVORED? If
#Q_(sp) > K_(sp)# , then yes... - Is it REACTANT-FAVORED? If
#Q_(sp) < K_(sp)# , then yes...

#Q_(sp) = ["Ag"^(+)]_i["Cl"^(-)]_i#

#= (9.77 xx 10^(-9) "M")("0.0100 M")#

#= 9.77 xx 10^(-11)#

And since **NOT** form a precipitate.

*The reaction favors the solid at the moment, so Le Chatelier's Principle suggests the reaction shifts to the right to dissolve more solid BEFORE precipitate is allowed to form, thus not allowing it to form.*