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.

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