Question

50.36 mL 2.74 ✕ 10⁻²M Pb(NO₃)₂ is mixed with 25.08 mL 4.68 ✕ 10⁻⁵M NaBr. (molar solubility of PbBr₂ = 1.2 ✕ 10⁻²M)

Ksp= 6.912e-6

calculate the Qsp

Someone previously gave me the answer Qsp= 3.66e-10 but this is INNCORRECT

Answer 1

Lets find the concentration after mixing for Pb(NO3)2

Concentration after mixing = mol of component / (total volume)

M(Pb(NO3)2) after mixing = M(Pb(NO3)2*)V(Pb(NO3)2)/(total volume)

M(Pb(NO3)2) after mixing = 0.0274 M*50.36 mL/(50.36+25.08)mL

M(Pb(NO3)2) after mixing = 1.829*10^-2 M

Lets find the concentration after mixing for NaBr

Concentration after mixing = mol of component / (total volume)

M(NaBr) after mixing = M(NaBr*)V(NaBr)/(total volume)

M(NaBr) after mixing = 4.68*10^-5 M*25.08 mL/(25.08+50.36)mL

M(NaBr) after mixing = 1.556*10^-5 M

The salt dissolves as:

PbBr2 <----> Pb2+ + 2 Br-

Qsp = [Pb2+][Br-]^2

Qsp = (1.829*10^-2)*(2*1.556*10^-5)^2

Qsp = 1.771*10^-11

we have,

Ksp = 6.912*10^-6

Since Qsp is less than ksp, precipitate will not form