In a discussion,a study argued that calcium ions when asded to toothpaste would enhance the performance of the toothpaste by further strengthening teeth. Therefore, he suggested the addition of calcium hydrogen carbonate, calculate the solubility of calcium fluoride in a solution containing calcium hydrogen carbonate of concentration 1×10-3 mol dm3
To calculate the solubility of calcium fluoride (CaF2) in a solution containing calcium hydrogen carbonate (Ca(HCO3)2) of concentration 1×10-3 mol dm3, you need to consider the solubility product constant (Ksp) for calcium fluoride.
The solubility product constant (Ksp) is a measure of the solubility of a compound in a solution and is derived from the equilibrium expression of the dissolution reaction. For calcium fluoride (CaF2), the dissolution reaction can be represented as:
CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
The solubility product constant expression (Ksp) for this reaction is defined as the product of the concentrations of the dissociated ions raised to the power of their respective stoichiometric coefficients:
Ksp = [Ca2+][F-]^2
To calculate the solubility (S) of calcium fluoride, we need to know the concentration of calcium ions ([Ca2+]) provided by the calcium hydrogen carbonate solution.
Given that the concentration of the calcium hydrogen carbonate solution is 1×10-3 mol dm3, it implies that the concentration of calcium ions ([Ca2+]) is also 1×10-3 mol dm3.
Now, we can substitute the concentrations into the solubility product constant expression:
Ksp = [Ca2+][F-]^2
Substituting [Ca2+] = 1×10-3 mol dm3, we get:
Ksp = (1×10-3 mol dm3)(S)^2
Finally, we can rearrange the equation to solve for the solubility (S) of calcium fluoride:
S = √(Ksp / [Ca2+])
So, find the Ksp value for calcium fluoride and substitute it along with the concentration of calcium ions to calculate the solubility of calcium fluoride in the given solution.