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
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
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.