When 20.0mL of 1.40M solution of calcium nitrate is mixed with 70.0mL of 0.234M potassium iodate, a precipitate Ca (IO3)2 is formed.
Calculate the concentrations of both NO3- and IO3- before mixing and after mixing.
(NO3^-) = 1.40 M x [2 nitrates/mol Ca(NO3)2] = ? before mixing.
(NO3^-) = 1.40 M x (20/90) x [2 nitrates/mol Ca(NO3)2] = ? after mixing. Remember nitrate ion is not involved in the rxn; it is a spectator ion.
(IO3^-) before mixing = 0.234 M.
After mixing is a lot of work to do it right and I don't know if your prof intended it to be done this way. Here are the two ways to do it; one long and one short. First, this is a limiting reagent (LR) problem with KIO3 being the LR. You have 20 x 1.4 = 28 mmols Ca(NO3)2 and 70 x 0.234 = 16.38 mmols KIO3 initially.
....Ca(NO3)2 + 2KIO3 =>Ca(IO3)2(s) + 2KNO3
I....28..........0.......0..........0
add............16.38.................
C..-16.38.....-16.38......8.19.........
E...11.62.........0.......8.19 solid.....
The easy, but incorrect way, of doing this is to say that IO3^- is zero from above BUT the Ca(IO3)3 ppt dissolves slightly and the solubility depends upon the concn of excess Ca^2+ since not all of that is used. Look up Ksp for Ca(IO3)2
Ksp = (Ca^2+)(IO3^-)^2
(Ca^2+) from above is 11.62 mmols/90 mL = ?
Substitute and solve for (IO3^-) and that is the (IO3^-) after mixing. Post any work if you have questions.
Thank you
To calculate the concentrations of NO3- and IO3- before and after mixing, we can use the principles of stoichiometry and the concept of dilution.
Before mixing:
1. Calculate the number of moles of calcium nitrate (Ca(NO3)2) in 20.0 mL of 1.40M solution:
molarity (M) = moles (mol) / volume (L)
1.40 mol/L = moles / 0.0200 L
Rearranging the equation to solve for moles:
moles = molarity * volume
moles = 1.40 mol/L * 0.0200 L
moles = 0.028 mol
Since calcium nitrate dissociates into two nitrate ions (NO3-), the concentration of NO3- can be calculated:
concentration (M) = moles (mol) / volume (L)
concentration of NO3- = 0.028 mol / 0.020 L
concentration of NO3- = 1.40 M
Similarly, the concentration of IO3- can be calculated using the same principles for potassium iodate (KIO3) solution:
concentration of IO3- = 0.234 M
After mixing:
Since a precipitate Ca(IO3)2 is formed, it means that calcium nitrate and potassium iodate reacted in a 1:2 molar ratio. This means that for every 1 mol of calcium nitrate, 2 mol of iodate ions are consumed.
Using this stoichiometric ratio, we can calculate the number of moles of iodate ions (IO3-) consumed:
moles of IO3- consumed = 2 * moles of Ca(NO3)2
moles of IO3- consumed = 2 * 0.028 mol
moles of IO3- consumed = 0.056 mol
Since the total volume after mixing is 20.0 mL + 70.0 mL = 90.0 mL, we need to convert it to liters:
total volume (L) = 90.0 mL / 1000 mL/L
total volume (L) = 0.0900 L
Now, we can calculate the concentrations of NO3- and IO3- after mixing:
concentration of NO3- = moles of NO3- / total volume (L)
concentration of NO3- = 0.028 mol / 0.0900 L
concentration of NO3- = 0.311 M
concentration of IO3- = moles of IO3- / total volume (L)
concentration of IO3- = 0.056 mol / 0.0900 L
concentration of IO3- = 0.622 M
Therefore, the concentrations of NO3- and IO3- before mixing are 1.40 M and 0.234 M, respectively, while after mixing, the concentrations are 0.311 M and 0.622 M, respectively.