'24.6 mL of 0.488 M NaCl is added to 24.6 mL of 0.312 M AgNO3. How many moles of AgCl would precipitate? What would be the concentrations of each of the ions in the reaction mixture after the reaction? Ion Concentration (M) Ag+ M NO3- M Na+ M Cl- M'
millimols NaCl = mL x M = 24.6 x 0.488 = estimated 12
mmols AgNO3 = estd 7.7
You need to redo these more accurately as well as all of the other numbers below. How advanced is this class? Basically, you have this but the final results depend upon just how complicated you want to make it. Remember these numbers are estimates.
..........NaCl + AgNO3 ==> AgCl + NaNO3
I.........12......7.7.......0........0
C........-7.7....-7.7......7.7.....
E.........4.3.....0?........7.7(solid)..
So Na^+ is a spectator ion. It is 12 initially and 12 at the end. M = mmols/total mL where total mL = 24.6+24.6 = ?
Cl was 12 and reduced by 7.7 = 4.3 left. M = mmols/mL
NO3^- is a spectator ion. 7.7 initially and 7.7 at the end. M = mmols/mL.
The (Ag^+) depends upon how advanced this calss is. If low level then (Ag^+) = 0 since all of it was pptd.
If the class is somewhat advanced the Ag^+ is what is left due to the solubility of AgCl in an excess of Cl^-
That is Ksp = (Ag^+)(Cl^-). Plug in Ksp from tables in your text/notes, plug in Cl from above and solve for Ag^+. Post your work if you get stuck.
To determine the number of moles of AgCl that would precipitate, we need to use stoichiometry. The balanced chemical equation for the reaction between NaCl and AgNO3 is as follows:
NaCl(aq) + AgNO3(aq) -> AgCl(s) + NaNO3(aq)
From the equation, we can see that 1 mole of AgCl is formed for every 1 mole of AgNO3 reacted.
First, we calculate the moles of NaCl and AgNO3:
Moles of NaCl = volume of NaCl (in L) x concentration of NaCl
= 0.0246 L x 0.488 M
= 0.0119968 moles
Moles of AgNO3 = volume of AgNO3 (in L) x concentration of AgNO3
= 0.0246 L x 0.312 M
= 0.0076632 moles
Since NaCl and AgNO3 react in a 1:1 ratio, the moles of AgCl formed will be equal to the moles of AgNO3 reacted.
Moles of AgCl = Moles of AgNO3
= 0.0076632 moles
Therefore, 0.0076632 moles of AgCl will precipitate.
To determine the concentrations of each ion in the reaction mixture after the reaction, we need to consider the volume of the final solution.
The total volume of the solution after the reaction is the sum of the initial volumes of the NaCl and AgNO3 solutions, which is 24.6 mL + 24.6 mL = 0.0492 L.
The moles of ions present in the final solution can be calculated using the following equations:
Moles of Na+ = moles of NaCl
= 0.0119968 moles
Moles of Ag+ = moles of AgNO3 - moles of AgCl
= 0.0076632 moles - 0.0076632 moles
= 0 moles
Moles of Cl- = moles of NaCl + moles of AgCl
= 0.0119968 moles + 0.0076632 moles
= 0.01966 moles
Moles of NO3- = moles of AgNO3
= 0.0076632 moles
Finally, we can calculate the concentrations of each ion:
Concentration of Na+ = moles of Na+ / total volume
= 0.0119968 moles / 0.0492 L
= 0.244 M
Concentration of Ag+ = moles of Ag+ / total volume
= 0 moles / 0.0492 L
= 0 M
Concentration of Cl- = moles of Cl- / total volume
= 0.01966 moles / 0.0492 L
= 0.4 M
Concentration of NO3- = moles of NO3- / total volume
= 0.0076632 moles / 0.0492 L
= 0.156 M
Therefore, the concentrations of the ions in the reaction mixture after the reaction are:
Ag+: 0 M
NO3-: 0.156 M
Na+: 0.244 M
Cl-: 0.4 M