Solve for
the number of grams of Ca2+
ion in a solution prepared by adding 1.00 gram of calcium oxalate (Ksp=1.3x10^-8)powder to 100 mL at pH 9.
I'm assuming that at a pH of 9 the H^+ will not increase the solubility over that of a saturated solution of CaC2O4.
............CaC2O4 ==> Ca^2+ + C2O4^2-
I............solid......0........0
C............solid......x........x
E............solid......x........x
Ksp = (Ca^2+)(C2O4^2-)
Substitue the E line and solve for x = (Ca^2+) in mols/L. Correct for mols in 100 mL and convert to grams.
To solve for the number of grams of Ca^2+ ion in the solution, we need to consider the dissociation of calcium oxalate and the solubility product constant.
First, let's write the balanced chemical equation for the dissociation of calcium oxalate:
CaC2O4 (s) ⇌ Ca^2+ (aq) + C2O4^2- (aq)
According to the stoichiometry of the equation, 1 mole of calcium oxalate (CaC2O4) will produce 1 mole of calcium ions (Ca^2+).
Now, we need to find the concentration of the calcium ions in the solution. Given that the solution has a pH of 9, we can assume that it is basic. In a basic solution, calcium ions (Ca^2+) will react with hydroxide ions (OH^-) to form calcium hydroxide (Ca(OH)2).
Ca^2+ (aq) + 2OH^- (aq) → Ca(OH)2 (s)
The reaction between Ca^2+ and OH^- indicates that OH^- ions are being consumed, gradually shifting the equilibrium to the right, causing the dissolution of the calcium hydroxide.
To calculate the concentration of the calcium ions, we need to determine the amount of OH^- ions consumed, which can be found from the pH of the solution. The concentration of OH^- ions in a basic solution can be obtained using the following formula:
[OH^-] = 10^(-pOH)
In this case, the pH of the solution is 9, so the pOH is 14 - 9 = 5. Therefore:
[OH^-] = 10^(-5)
Next, let's consider the reaction between calcium hydroxide and calcium oxalate. Calcium hydroxide reacts with calcium oxalate to form insoluble calcium oxalate:
Ca(OH)2 (s) + CaC2O4 (s) → CaC2O4 (s) + 2OH^- (aq)
Since calcium oxalate is insoluble, it will remain as a solid throughout the reaction. Thus, the formation of calcium oxalate does not consume OH^- ions.
Now, let's consider the solubility product constant (Ksp) for calcium oxalate. The Ksp expression for the equilibrium can be written as:
Ksp = [Ca^2+][C2O4^2-]
Since the concentration of [C2O4^2-] can be considered negligible (as calcium oxalate is solid), we can assume that [C2O4^2-] is close to zero. Therefore, we can simplify the equation to:
Ksp = [Ca^2+]
Now, we can substitute the Ksp value given in the problem, and we get:
1.3 x 10^(-8) = [Ca^2+]
Finally, multiplying the concentration of calcium ions [Ca^2+] by the volume of the solution (100 mL) and then converting to grams will give us the number of grams of Ca^2+ ion in the solution.
Note: To convert mL to L, divide by 1000:
100 mL ÷ 1000 = 0.1 L
So, the number of grams of Ca^2+ ion in the solution is calculated as follows:
[Ca^2+] in mol/L = 1.3 x 10^(-8) mol/L
[Ca^2+] in g/L = (1.3 x 10^(-8) mol/L) * (40.08 g/mol) = 5.2 x 10^(-7) g/L
Number of grams of Ca^2+ = [Ca^2+] * Volume of solution
Number of grams of Ca^2+ = (5.2 x 10^(-7) g/L) * (0.1 L) = 5.2 x 10^(-8) g