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.
I believe this is a screwed up question from the get go.
To determine the grams of potassium iodate needed to precipitate calcium iodate, we need to consider the solubility product constant (Ksp) and the molarity of the calcium iodate solution.
A) To begin precipitating calcium iodate from the solution, we need to reach a concentration of calcium and iodate ions that exceeds the Ksp value. The balanced chemical equation for the precipitation reaction is:
Ca(IO3)2 (aq) ⇌ Ca(IO3)2 (s)
According to the equation, the molar ratio between calcium iodate and iodate ions is 1:2. Thus, for every mole of calcium iodate formed, 2 moles of iodate ions are required.
To calculate the number of moles of iodate ions present in the 250 ml solution of 0.13M calcium iodate, we multiply the molarity by the volume:
Moles of iodate ions = Molarity × Volume
= 0.13 M × 0.25 L
= 0.0325 moles
Since the molar ratio between calcium iodate and iodate ions is 1:2, we need half the number of moles of calcium iodate:
Moles of calcium iodate = 0.0325 moles ÷ 2
= 0.01625 moles
To convert this number of moles to grams, we need to know the molar mass of calcium iodate. Calcium (Ca) has a molar mass of approximately 40.08 g/mol, iodine (I) has a molar mass of approximately 126.90 g/mol, and oxygen (O) has a molar mass of approximately 16.00 g/mol. Therefore, the molar mass of calcium iodate (Ca(IO3)2) is:
Molar mass of calcium iodate = (40.08 g/mol) + 2 × [(126.90 g/mol) + 3 × (16.00 g/mol)]
= 40.08 g/mol + 2 × (126.90 g/mol + 48.00 g/mol)
= 40.08 g/mol + 2 × (174.90 g/mol)
= 40.08 g/mol + 349.80 g/mol
= 389.88 g/mol
Now, we can calculate the grams of calcium iodate needed by multiplying the moles of calcium iodate by its molar mass:
Grams of calcium iodate = Moles of calcium iodate × Molar mass of calcium iodate
= 0.01625 moles × 389.88 g/mol
≈ 6.343 grams
Therefore, approximately 6.343 grams of potassium iodate need to be added to the solution to begin precipitating calcium iodate.
B) To find the total grams of potassium iodate needed to saturate the solution, we need to consider the Ksp value. The balanced chemical equation for the dissolution of potassium iodate (KIO3) is:
KIO3 (s) ⇌ K+ (aq) + IO3- (aq)
According to the equation, for every mole of potassium iodate that dissolves, one mole of potassium ions and one mole of iodate ions are produced. The molar ratio between potassium iodate and iodate ions is 1:1.
To calculate the total number of moles of iodate ions required to reach saturation, we need to divide the Ksp value by the molar concentration:
Moles of iodate ions = Ksp ÷ Molarity
= 0.183 ÷ 0.13 M
≈ 1.408 moles
Since the molar ratio between potassium iodate and iodate ions is 1:1, we need the same number of moles of potassium iodate.
To convert this number of moles to grams, we multiply the moles by the molar mass of potassium iodate. Potassium (K) has a molar mass of approximately 39.10 g/mol, iodine (I) has a molar mass of approximately 126.90 g/mol, and oxygen (O) has a molar mass of approximately 16.00 g/mol. Therefore, the molar mass of potassium iodate (KIO3) is:
Molar mass of potassium iodate = (39.10 g/mol) + (126.90 g/mol) + 3 × (16.00 g/mol)
= 39.10 g/mol + 126.90 g/mol + 48.00 g/mol
= 39.10 g/mol + 174.90 g/mol
= 214.00 g/mol
Now, we can calculate the grams of potassium iodate needed by multiplying the moles of potassium iodate by its molar mass:
Grams of potassium iodate = Moles of potassium iodate × Molar mass of potassium iodate
= 1.408 moles × 214.00 g/mol
≈ 301.712 grams
Therefore, approximately 301.712 grams of potassium iodate need to be added to the solution until it is saturated in potassium iodate.