Respuesta :
The experimental value for Ksp of Ca(OH)2 obtained was much higher than the literature value,
% error = (4.50 x 10^-3 - 4.68 x 10^-6) x 100/4.50 x 10^-3 = 99.90%
The error could have occured due to higher dilution of the solution which enhances the solubility of Ca(OH)2 in the solution.
% error = (4.50 x 10^-3 - 4.68 x 10^-6) x 100/4.50 x 10^-3 = 99.90%
The error could have occured due to higher dilution of the solution which enhances the solubility of Ca(OH)2 in the solution.
The solubility equilibrium is the constant for the solid solute that gets dissolved in an aqueous solution. The experimental value is higher than the literature value.
What is solubility equilibrium?
Solubility or the equilibrium constant (Ksp) is the level at which the solute gets dissolved in a solution and depicts the solubility of the solute.
The percentage error is calculated as:
[tex]\dfrac{(4.50 \times 10^{-3} - 4.68 \times 10^{-6}) \times 100}{4.50\times 10^{-3}} = 99.90\%[/tex]
The higher dilution or unsaturated calcium hydroxide may be the reason for the error and increased solubility of the solute.
Therefore, the experimental is higher than the theoretical value.
Learn more about solubility equilibrium here:
https://brainly.com/question/15241577
