Question

When PbBr2(s) is added to 2.5 L of water, what mass of PbBr2 will dissolve? Ksp (PbBr2) = 4.6 x 10-6. Group of answer choices: A. 0.010g B.9.6 g C. 12.1g D. 0.56g E. 3.8 g *** please show work and explain!!! Thank you

Answer 1

At equilibrium:
PbBr2     <---->     Pb2+     +         2 Br-   

                     s                  2s      

Ksp = [Pb2+][Br-]^2
4.6*10^-6=(s)*(2s)^2
4.6*10^-6= 4(s)^3
s = 1.048*10^-2 M

Molar mass of PbBr2,
MM = 1*MM(Pb) + 2*MM(Br)
= 1*207.2 + 2*79.9
= 367 g/mol


Molar mass of PbBr2= 367.0 g/mol
s = 1.048*10^-2 mol/L
To covert it to g/L, multiply it by molar mass
s = 1.048*10^-2 mol/L * 367.0 g/mol
s = 3.845 g/L

Here,
Volume is 2.5 L
So,
Mass = s * volume
= 3.845 g/L * 2.5 L
= 9.6 g
Answer: B