If 100. mL of 0.400 M Na2SO4 is added to 200. mL of 0.600 M NaCl, what is the concentration of Na+ ions in the final solution? Assume that the volumes are additive.
moles Na2SO4 = M x L.
moles Na^+ = 2x that.
moles NaCl = M x L
moles Na^+ = 1x that.
M Na^+ = total moles Na^+/total volume in L.
Thank You got the right answer
Oh, I see you're mixing up some salty business here! Let's calculate the concentration of Na+ ions in the final solution.
First, we need to find the moles of Na+ ions in each solution. In the 100 mL of 0.400 M Na2SO4 solution, we have:
(0.400 mol/L) * (0.100 L) = 0.040 mol of Na+ ions
Similarly, in the 200 mL of 0.600 M NaCl solution, we have:
(0.600 mol/L) * (0.200 L) = 0.120 mol of Na+ ions
Now, we add the moles of Na+ ions from both solutions:
0.040 mol + 0.120 mol = 0.160 mol
To find the concentration of Na+ ions in the final solution, we divide the total moles by the total volume:
0.160 mol / (100 mL + 200 mL) = 0.160 mol / 0.300 L = 0.533 M
So, the concentration of Na+ ions in the final solution is 0.533 M. Keep those ions mixed and the sodium party going!
To find the concentration of Na+ ions in the final solution, we need to determine the amount of Na+ ions present after mixing the two solutions.
First, let's find the amount of Na+ ions in each solution separately, using the formula:
amount (in moles) = concentration (in M) x volume (in L)
For the Na2SO4 solution:
Amount of Na+ ions = concentration of Na2SO4 x volume of Na2SO4
Amount of Na+ ions = 0.400 M x 0.100 L = 0.040 moles
For the NaCl solution:
Amount of Na+ ions = concentration of NaCl x volume of NaCl
Amount of Na+ ions = 0.600 M x 0.200 L = 0.120 moles
Now, let's add the amounts of Na+ ions from both solutions together:
Total amount of Na+ ions = amount of Na+ ions from Na2SO4 + amount of Na+ ions from NaCl
Total amount of Na+ ions = 0.040 moles + 0.120 moles = 0.160 moles
Finally, to find the concentration of Na+ ions in the final solution, divide the total amount of Na+ ions by the total volume of the solution.
Total volume of the final solution = volume of Na2SO4 + volume of NaCl
Total volume of the final solution = 0.100 L + 0.200 L = 0.300 L
Concentration of Na+ ions in the final solution = Total amount of Na+ ions / Total volume of the final solution
Concentration of Na+ ions in the final solution = 0.160 moles / 0.300 L = 0.533 M
Therefore, the concentration of Na+ ions in the final solution is 0.533 M.