You have a 250ml solution of 0.13M calcium iodate. How many grams of potassium iodate need to be added to the solution to begin to precipitate calcium iodate? The solubility product constant for calcium iodate is6.47 *10^-6.
B) you have a 250ml solution of 0.13 M calcium iodate. How many total grams need to be added to the solution until it is saturated in potassium iodate? The ksp for potassium Iodate is 0.183.
pls help me with this problem
Neither problem makes any sense to me. With Ksp = 6.47E-6, the solubility of Ca(IO3)2 is 0.0117 in a saturated solution. Therefore, you can't have a 0.13M solution. You don't need to add any KIO3; Ca(IO3)2 is already pptd.
Sure! Let's break down the two parts of the problem separately:
Part A: Precipitation of Calcium Iodate
To determine how many grams of potassium iodate need to be added to the solution to begin precipitating calcium iodate, we need to calculate the maximum amount of calcium iodate that can be dissolved in the 250 mL solution, based on the solubility product constant (Ksp) for calcium iodate.
The Ksp expression for calcium iodate (Ca(IO3)2) is written as:
Ksp = [Ca2+][IO3-]^2
Given that the solubility product constant (Ksp) for calcium iodate is 6.47 × 10^-6, and considering the stoichiometry of the compound, we can assume that each calcium iodate molecule dissociates into one calcium ion (Ca2+) and two iodate ions (IO3-).
Let's assume 'x' moles of calcium iodate is dissolved in the solution. Therefore, the concentration of calcium ion and iodate ion will be 'x mol/L' and '2x mol/L', respectively.
Hence, the Ksp expression can be written as:
Ksp = (x)(2x)^2 = 6.47 × 10^-6
Simplifying the equation, we get:
4x^3 = 6.47 × 10^-6
x^3 = (6.47 × 10^-6) / 4
x^3 = 1.6175 × 10^-6
Taking the cube root of both sides of the equation:
x = ∛(1.6175 × 10^-6)
x ≈ 0.012 mol/L
Now, since we know the volume of the solution is 250 mL, we can calculate the number of moles of calcium iodate that can be dissolved in the solution using the concentration and volume relationship:
Number of moles of calcium iodate = concentration × volume
= 0.012 mol/L × 0.25 L
= 0.003 mol
To calculate the mass of calcium iodate, we can use the molar mass of calcium iodate, which can be calculated by adding the atomic masses of calcium (Ca), iodine (I), and oxygen (O) together:
Ca(IO3)2 = (40.08 g/mol) + (2 × 126.90 g/mol) + (6 × 16.00 g/mol)
= 40.08 g/mol + 253.80 g/mol + 96.00 g/mol
= 389.88 g/mol
Mass of calcium iodate = number of moles × molar mass
= 0.003 mol × 389.88 g/mol
≈ 1.17 grams
Therefore, approximately 1.17 grams of potassium iodate need to be added to the solution to begin precipitating calcium iodate.
Part B: Saturation of Potassium Iodate
In this part, we need to calculate the total amount of potassium iodate needed to saturate the 250 mL solution, based on the solubility product constant (Ksp) for potassium iodate.
The Ksp expression for potassium iodate (KIO3) is written as:
Ksp = [K+][IO3-]
Given that the solubility product constant (Ksp) for potassium iodate is 0.183, and considering the stoichiometry of the compound, we can assume that each potassium iodate molecule dissociates into one potassium ion (K+) and one iodate ion (IO3-).
Let's assume 'y' moles of potassium iodate is dissolved in the solution. Therefore, the concentration of potassium ion and iodate ion will be 'y mol/L' each.
Hence, the Ksp expression can be written as:
Ksp = (y)(y) = 0.183
Simplifying the equation, we get:
y^2 = 0.183
y ≈ √0.183
y ≈ 0.428 mol/L
Now, since we know the volume of the solution is 250 mL, we can calculate the number of moles of potassium iodate that can be dissolved in the solution using the concentration and volume relationship:
Number of moles of potassium iodate = concentration × volume
= 0.428 mol/L × 0.25 L
= 0.107 mol
To calculate the mass of potassium iodate, we can use the molar mass of potassium iodate, which can be calculated by adding the atomic masses of potassium (K), iodine (I), and oxygen (O) together:
KIO3 = 39.10 g/mol + 126.90 g/mol + (3 × 16.00 g/mol)
= 166.10 g/mol
Mass of potassium iodate = number of moles × molar mass
= 0.107 mol × 166.10 g/mol
≈ 17.78 grams
Therefore, approximately 17.78 grams of potassium iodate need to be added to the solution until it is saturated in potassium iodate.
I hope this helps! Let me know if you have any further questions.