Asked by Michelle
At 25 °C, only 1.9 g CaSO4 (MM = 136 g/mol) will dissolve in 2.00 L of water. Write the equilibrium
expression and calculate the equilibrium constant for the reaction below.
CaSO4 (s) ⇋ Ca2+ (aq) + SO42- (aq)
expression and calculate the equilibrium constant for the reaction below.
CaSO4 (s) ⇋ Ca2+ (aq) + SO42- (aq)
Answers
Answered by
DrBob222
Ksp = (Ca^+2)(SO4^-2)
Solubility CaSO4 = S = 1.9g/136 = ??mole and that in 2L is ??/2 = (S)
Then substitute the value for S into the Ka expression and solve for Ka.
Solubility CaSO4 = S = 1.9g/136 = ??mole and that in 2L is ??/2 = (S)
Then substitute the value for S into the Ka expression and solve for Ka.
Answered by
Galaxy Dalton
nothing
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