Answer :
Pbcl2 precipitate when 275 ml of a 0.134 m solution of pb(no3)2 is added to 125 ml of 0.0339 m solution of nacl? the ksp of pbcl2 is 1.17e-5.
Given that,
The volume of NaCl = 125 ml
Concentration of NaCl = 0.0339 m
The volume of Pb(NO3)2 = 275 ml
Concentration of Pb(NO3)2 = 0.134 m
K(eq) = 1.17 × 10(-5) --------(1)
Total volume = 275 + 125 = 400 ml
Pb(NO3)2 ------- Pb(+2) + 2 NO3(-)
at t = 0 0.134 0 0
at t = t 0 0.134 0.134
NaCl -------- Na(+) + Cl(+)
at t = 0 0. 0339 0 0
at t = t 0 0.0339 0.0339
PbCl2 ---------- Pb(+2) + 2Cl(-)
Q(sp) = [Pb+2] [ Cl-]^2
By substituting the values of conc. of all the ions, we get Pbcl2 precipitate when 275 ml of a 0.134 m solution of pb(no3)2 is added to 125 ml of 0.0339 m solution of nacl? the ksp of pbcl2 is 1.17e-5.
= (275 × 0.134/ 400) (125 × 0.0339/400)^2
= 9.8 × 10(-6)
As we noticed that, Ksp > Qsp
Therefore, PbCl2 will not precipitate.
Thus, we concluded that the Pbcl2 precipitate when 275 ml of a 0.134 m solution of pb(no3)2 is added to 125 ml of 0.0339 m solution of nacl? the ksp of pbcl2 is 1.17e-5.
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