You have 3.00 L of a 2.37 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 M solution of AgNO3(aq) called solution B. You mix these solutions together, making solution C.
Calculate the concentration (in M) of Na+ ions in solution C.
To find the concentration of Na+ ions in solution C, we need to determine the number of moles of Na+ ions present in the total volume of solution. This can be done using the formula:
moles = concentration (in M) × volume (in L)
First, let's calculate the number of moles of NaCl in solution A:
moles of NaCl = concentration of NaCl (in M) × volume of solution A (in L)
= 2.37 M × 3.00 L
= 7.11 mol
Since NaCl dissociates into one Na+ ion for every Cl- ion, the number of moles of Na+ ions in the solution A is also 7.11 mol.
Next, let's calculate the number of moles of NaNO3 in solution B:
moles of NaNO3 = concentration of NaNO3 (in M) × volume of solution B (in L)
= 2.00 M × 2.00 L
= 4.00 mol
Again, since NaNO3 dissociates into one Na+ ion for every NO3- ion, the number of moles of Na+ ions in solution B is 4.00 mol.
Now, when we mix these solutions together, the total volume of the resulting solution is:
total volume = volume of solution A + volume of solution B
= 3.00 L + 2.00 L
= 5.00 L
The total number of moles of Na+ ions in the resulting solution C is the sum of the moles of Na+ ions from solution A and solution B:
total moles of Na+ = moles of Na+ ions from solution A + moles of Na+ ions from solution B
= 7.11 mol + 4.00 mol
= 11.11 mol
Finally, we can calculate the concentration of Na+ ions in solution C by dividing the total number of moles of Na+ ions by the total volume of the solution:
concentration of Na+ in solution C = total moles of Na+ ions / total volume of solution C
= 11.11 mol / 5.00 L
= 2.22 M
Therefore, the concentration of Na+ ions in solution C is 2.22 M.
To calculate the concentration of Na+ ions in solution C, we need to determine the number of moles of Na+ ions present in the solution and then divide it by the total volume of the solution.
First, let's find the number of moles of NaCl in solution A:
Number of moles of NaCl = concentration (M) × volume (L)
Number of moles of NaCl = 2.37 M × 3.00 L
Next, we need to find the number of moles of Na+ ions in that amount of NaCl since NaCl dissociates into Na+ and Cl- ions in solution. We know that 1 mole of NaCl produces 1 mole of Na+ ions. Therefore, the number of moles of Na+ ions in solution A is also 2.37 M × 3.00 L.
Similarly, let's find the number of moles of Na+ ions in solution B:
Number of moles of Na+ ions in solution B = concentration (M) × volume (L)
Number of moles of Na+ ions in solution B = 2.00 M × 2.00 L
Now, we can find the total number of moles of Na+ ions by adding the moles from both solutions A and B:
Total number of moles of Na+ ions = (2.37 M × 3.00 L) + (2.00 M × 2.00 L)
To calculate the concentration of Na+ ions in solution C, we divide the total number of moles of Na+ ions by the total volume of the solution (which is the sum of the volumes of solutions A and B):
Concentration of Na+ ions in solution C = Total moles of Na+ ions / Total volume of solution C
Substituting the values we previously calculated, we can now determine the concentration of Na+ ions in solution C.
The sodium ion is a spectator ion in this reaction, i.e. it does not take part. Thus we can ignore the chemistry.
Number of moles at start
3.00 L x 2.37 mole L^-1
Because 1 mole of NaCl gives 1 mole of Na+ in solution.
volume at end = 3.00 L + 2.00 L
=5.00 L
so concentration of Na+ at end is
moles/volume
=3.00 L x 2.37 mole L^-1/5.00 L