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 to solve this problem
To answer these questions, we need to consider the concept of solubility and the solubility product constant (Ksp).
A) To determine the grams of potassium iodate needed to precipitate calcium iodate from the solution, we first need to calculate the concentration of calcium iodate ions in the solution using the given molarity (0.13 M) and volume (250 ml).
Step 1: Convert the volume to liters:
250 ml = 0.250 L
Step 2: Calculate the number of moles of calcium iodate in the solution:
mol = molarity x volume
mol = 0.13 mol/L x 0.250 L
Step 3: Use the balanced equation for the reaction of calcium iodate forming a precipitate:
Ca(IO3)2 (aq) ⇌ Ca(IO3)2 (s)
Since the equation shows that 1 mole of calcium iodate (Ca(IO3)2) forms 1 mole of calcium iodate precipitate, the molar ratio is 1:1.
Therefore, the number of moles of calcium iodate precipitate formed is equal to the number of moles of calcium iodate in solution.
Step 4: Use the solubility product constant (Ksp) to calculate the concentration of calcium iodate ions (Ca(IO3)2) in the solution when it begins to precipitate.
Ca(IO3)2 ⇌ Ca2+ + 2IO3-
Ksp = [Ca2+][IO3-]^2 = 6.47 x 10^-6
We know the concentration of Ca2+ is equal to the initial concentration of calcium iodate, which is 0.13 M.
Plugging in the values, we have:
6.47 x 10^-6 = (0.13 M)(x)^2
Solving for x (concentration of IO3-), we find:
x = sqrt(6.47 x 10^-6 / 0.13)
Now we know the concentration of IO3- ions when calcium iodate begins to precipitate.
Step 5: Convert the concentration back to moles:
mol = concentration x volume
mol = x (from step 4) x 0.250 L
Step 6: Convert moles to grams:
To convert moles to grams, we need to know the molar mass of potassium iodate (KIO3):
K: 39.10 g/mol
I: 126.90 g/mol
O: 16.00 g/mol
molar mass of KIO3 = (39.10 g/mol) + (126.90 g/mol) + (16.00 g/mol x 3) = 214.00 g/mol
grams = mol x molar mass
grams = moles x 214.00 g/mol
So, the grams of potassium iodate needed to begin precipitating calcium iodate are given by the equation:
grams = (x from step 4) x 0.250 L x 214.00 g/mol
B) To determine the total grams of potassium iodate needed to saturate the solution, we need to consider the solubility product constant (Ksp) for potassium iodate (KIO3), which is given as 0.183.
Step 1: Calculate the concentration of KIO3 ions in solution when it is saturated using the Ksp:
KIO3 ⇌ K+ + IO3-
Ksp = [K+][IO3-] = 0.183
We know the concentration of K+ is equal to the initial concentration of potassium iodate (0.13 M). Therefore, the concentration of IO3- can be calculated using the Ksp equation.
Step 2: Convert the concentration of IO3- to moles using the given volume (250 ml), and convert moles to grams as in Step 6 of the previous case.
This will give you the total grams of potassium iodate needed to saturate the solution.
Note: It is important to consider that these calculations assume ideal behavior and do not account for factors such as temperature or other possible interactions occurring in the solution.