Given #K_"sp"# #ZnCO_3=1.46xx10^-10#, what is the #"ppm"# concentration of a saturated solution?

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Answer 1 · 2017-07-05T02:12:53

#S = 1.21xx10^-5M = 1.51xx10^-3(g/L)#

#ZnCO_3 rightleftharpoons Zn^(+2) + CO_3^(2-)# is a 1:1 ionization ratio. The solubility can be determined using #S = sqrt(Ksp)#

#S = sqrt(Ksp) = sqrt(1.46xx10^-10)M = 1.21xx10^-5M#

#1.21xx10^-5(mol)/Lxx (125g)/"mol" = 1.51xx10^-3(g/L)#

Answer 2 · 2017-07-05T03:51:46

#"Solubility"_(ZnCO_3) = "1 ppm"#

We consider the equilibrium reaction.........

#ZnCO_3(s) stackrel(H_2O)rightleftharpoonsZn^(2+) + CO_3^(2-)#

And #K_(sp)=[Zn^(2+)][CO_3^(2-)]=1.46xx10^-10#

We let the #"Solubility"_(ZnCO_3) = x*mol*L^-1#

And thus #"Solubility"_(ZnCO_3)=sqrt(1.46xx10^-10)#

#=1.21xx10^-5*mol*L^-1#

And thus the #"gram solubility"# of #"zinc carbonate"#

#-=1.21xx10^-5*mol*L^-1xx125.4*g*mol^-1~=1*mg*L^-1#, i.e. #"ppm"# concentrations.