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
Aw c'mon. It takes time to work these things and type them up. See your post below and please be patient.
To calculate the concentrations of NO3- and IO3- before mixing and after mixing the two solutions, we can use the concept of moles and the principle of conservation of mass.
Let's start by calculating the number of moles of NO3- and IO3- in the separate solutions before mixing:
For calcium nitrate solution:
Given:
Volume (V1) = 20.0 mL = 0.0200 L
Concentration (C1) = 1.40 M
Number of moles (n1) = V1 * C1
= 0.0200 L * 1.40 M
= 0.028 moles
For potassium iodate solution:
Given:
Volume (V2) = 70.0 mL = 0.0700 L
Concentration (C2) = 0.234 M
Number of moles (n2) = V2 * C2
= 0.0700 L * 0.234 M
= 0.0164 moles
So, before mixing, the number of moles of NO3- in the calcium nitrate solution is 0.028 moles, and the number of moles of IO3- in the potassium iodate solution is 0.0164 moles.
Now, let's calculate the number of moles of NO3- and IO3- after the two solutions are mixed:
Since calcium nitrate and potassium iodate react to form Ca(IO3)2, the reaction can be represented as:
Ca(NO3)2 + KIO3 -> Ca(IO3)2 + 2KNO3
From the balanced equation, we can see that for every molecule of Ca(NO3)2, one molecule of Ca(IO3)2 is formed.
Given that we have excess Ca(NO3)2 and the reaction goes to completion, the moles of Ca(IO3)2 formed will be equal to the moles of Ca(NO3)2 present in the calcium nitrate solution before mixing.
Therefore, the number of moles of Ca(IO3)2 formed = moles of Ca(NO3)2 = 0.028 moles.
Since 1 molecule of Ca(IO3)2 contains 1 mole of IO3-, the number of moles of IO3- formed will also be equal to 0.028 moles.
Thus, after mixing, the number of moles of NO3- remains the same at 0.028 moles, and the number of moles of IO3- is 0.028 moles.
To calculate the concentrations of NO3- and IO3- after mixing, we need to divide the number of moles by the total volume of the mixture.
Total volume (V_total) = V1 + V2 = 0.0200 L + 0.0700 L = 0.0900 L
Concentration of NO3- after mixing (C_NO3-) = moles of NO3- / V_total
= 0.028 moles / 0.0900 L
= 0.311 M
Concentration of IO3- after mixing (C_IO3-) = moles of IO3- / V_total
= 0.028 moles / 0.0900 L
= 0.311 M
Therefore, the concentrations of both NO3- and IO3- after mixing are 0.311 M.