Respuesta :

First balance the equation:
CaC2O4 (s) -> Ca2+ (aq) + C2O4 2- (aq)

from the balanced equation, write Ksp:
Ksp=[Ca2+][C2O4]  (Calcium oxalate is not included because it is a solid)

convert mg to g and then g/L into Molarity (Mol/L) to find the concentrations used in finding Ksp:
0.67mg/L x (1 g/ 1000 mg) x (1 mol/ 128.1 g) = 5.23e-6 M

Set up an ICE table from the balanced equation with previous calculation as the amount that has dissociated: (again, you can ignore Calcium oxalate because it is a solid and not included in Ksp)

    CaC2O4 (s) -> Ca2+ (aq) + C2O4 (aq)
I    -                      0                    0
C  -                       +5.23e-6       +5.23e-6
E  -                       +5.23e-6       +5.23e-6

You know have the concentrations of the species necessary for Ksp:
Ksp=(Ca2+)(C2O4)
=(5.23e-6)(5.23e-6)

Ksp = 2.74e-11
okpalawalter8

The Ksp(solubility products constant) is mathematically given as

Ksp = 2.74*10^-11

What is its Ksp?

Ksp:Ksp denotes solubility products constant, this is used to refer to the eqiulbrum constant that exist between a solid substance and its solution.

Question Parameters:

Calcium oxalate, cac2o4 (m = 128.1), dissolves to the extent of 0.67 mg l–1 .

Generally, the equation for the   is mathematically given as

CaC2O4 (s) -> Ca2+ (aq) + C2O4 2- (aq)

Where, the concentrations  in  Ksp is

0.67mg/L x (1 g/ 1000 mg) x (1 mol/ 128.1 ) = 5.23*10^-6 M

Therefore

Ksp=(Ca2+)(C2O4)

Ksp=(5.23e-6)(5.23*10^-6)

Ksp = 2.74*10^-11

In conclusion, the solubility products constant of the reaction is given as

Ksp = 2.74*10^-11

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