Discover the focus in moles per L, so:
1.56E-2 g/L (1 mol / 156.07 g CaCrO4) = 1E-4 mol / L = 1E-4 M CaCrO4
CaCrO4 <-> Ca CrO4
ksp = [Ca ][CrO4] =(1E-4)^2 = 1E-8
Reply 6
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RE:
The solubility of calcium chromate, CaCrO4, is 1.5610-2 g/L of resolution. Whats the Ksp for CaCrO4
Molar mass CaCrO4 = 156.0723 g/mol
1.5610^-2g = (1.5610^-2)/156.0723 = 1.0*10^-4 mol
Molarity of CaCrO4 = 1.0*10^-4M
CaCrO4 dissociates:
CaCrO4 Ca2 CrO4 2-
Ksp =[Ca2 ] *[ CrO4 2-]
Ksp = [1.04*10^-4]
Ksp = 1.0*10^-8
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Moles CaCrO4 = 1.56 x 10^-3 mol / 156.06 g/mol = 1.0 10^-6 in 100 mL 1.0 x 10^-6 in 1 L ( molar solubility) Ksp = 1.0 x 10^-6 ( 1.0 x 10^-6) =1.0 x 10^-11
Reply 6
Molar mass CaCrO4 = 156.0723 g/mol
1.5610^-2g = (1.5610^-2)/156.0723 = 1.0*10^-4 mol
Molarity of CaCrO4 = 1.0*10^-4M
CaCrO4 dissociates:
CaCrO4 Ca2 CrO4 2-
Ksp =[Ca2 ] *[ CrO4 2-]
Ksp = [1.04*10^-4]
Ksp = 1.0*10^-8
This post is last updated on hrtanswers.com at Date : 1st of September – 2022