A 1.00-L solution saturated at 25°C with calcium oxalate (CaC2O4) contains 0.0061 g of CaC2O4. The solubility constant (Ksp) for this salt is [CaC2O4=128.09 g/mol]
To solve this problem, we can use the solubility product constant (Ksp) equation:
CaC2O4(s) ⇋ Ca2+(aq) + C2O42-(aq)
Given:
- Volume of the solution (V) = 1.00 L
- Mass of CaC2O4 dissolved (m) = 0.0061 g
- Molar mass of CaC2O4 = 128.09 g/mol
Step 1: Calculate the concentration of Ca2+ and C2O42- ions in the solution.
Since CaC2O4 is a 1:1 salt, the concentration of Ca2+ ions and C2O42- ions will be the same. Let's assume both concentrations as x (mol/L).
CaC2O4(s) ⇋ Ca2+(aq) + C2O42-(aq)
Initial concentration: 0 M 0 M 0 M
Change in concentration: x M x M x M
Equilibrium concentration: x M x M x M
Step 2: Write the expression for the solubility product constant (Ksp).
The Ksp expression for this reaction is as follows:
Ksp = [Ca2+][C2O42-]
Ksp = x * x = x^2
Step 3: Calculate the value of x based on the given information.
Since the initial concentration of CaC2O4 is 0 M and all of it dissolves, the equilibrium concentration of Ca2+ and C2O42- ions will be the same, which is x.
Therefore, x = 0.0061 g / (128.09 g/mol) = 0.0000476 mol
Step 4: Substitute the value of x into the Ksp expression to find the solubility product constant (Ksp).
Ksp = (0.0000476 mol/L)^2
Ksp = 2.26 x 10^-9 mol^2/L^2
The solubility product constant (Ksp) for calcium oxalate (CaC2O4) is 2.26 x 10^-9 mol^2/L^2.
To find the value of the solubility constant (Ksp) for calcium oxalate (CaC2O4), we need to use the given information.
Step 1: Write the balanced equation for the dissociation of CaC2O4 into its ions:
CaC2O4 (s) ⇌ Ca^2+ (aq) + C2O4^2- (aq)
Step 2: Identify the stoichiometry of the balanced equation:
The balanced equation shows that one mole of CaC2O4 dissociates to form one mole of Ca^2+ ions and one mole of C2O4^2- ions.
Step 3: Calculate the molar concentration (concentration in mol/L or M) of Ca^2+ ions and C2O4^2- ions in the saturated solution.
To do this, we need to convert the given mass of CaC2O4 to moles and then divide by the volume of the solution.
Given:
Mass of CaC2O4 = 0.0061 g
Molar mass of CaC2O4 = 128.09 g/mol
Volume of solution = 1.00 L
Number of moles of CaC2O4 = Mass / Molar mass
= 0.0061 g / 128.09 g/mol
≈ 4.76 x 10^-5 mol
Since the stoichiometry of the balanced equation is 1:1, the concentration of Ca^2+ and C2O4^2- ions in the solution is equal to the number of moles of CaC2O4.
Concentration of Ca^2+ ions = 4.76 x 10^-5 mol/L
Concentration of C2O4^2- ions = 4.76 x 10^-5 mol/L
Step 4: Calculate the solubility product (Ksp) using the concentrations of the ions:
Ksp = [Ca^2+] * [C2O4^2-]
= (4.76 x 10^-5) * (4.76 x 10^-5)
= 2.27 x 10^-9
Therefore, the solubility constant (Ksp) for calcium oxalate (CaC2O4) is 2.27 x 10^-9.
0.0061g/128.09 = 4.76E-5
.........CaC2O4 ==> Ca^2+ + C2O4^2-
equil....4.76E-5..4.76E-5...4.76E-5
Ksp = (Ca^2+)(C2O4^2-)
Substitute from the ICE chart and solve for Ka.