Calculate the final concentrations in mol/L of H+, Na+, Cl- and SO42- when the following three solutions are mixed together:
1000 cm3 of 0.10 mol dm-3 HCl
500 cm3 of 0.20 mol dm-3 NaCl
500 cm3 of 0.20 mol dm-3 of Na2SO4
calculate moles of H+ added: 1.000*.1=.1Mol
calculate moles of Na: .5*.2+.5*.2*2
calculate moles of Cl : 1.0*.1+.5*.2
and so one with the sulfate ion.
Finally, divide each by the total volume, 2liters.
To calculate the final concentrations of H+, Na+, Cl-, and SO42- when the solutions are mixed together, we can use the principle of conservation of mass and the concept of moles.
First, let's determine the number of moles of each species present in the original solutions:
For the 1000 cm3 of 0.10 mol dm-3 HCl solution:
Number of moles of H+ = 0.10 mol dm-3 x 1000 cm3 / 1000 cm3 = 0.10 mol
For the 500 cm3 of 0.20 mol dm-3 NaCl solution:
Number of moles of Na+ = 0.20 mol dm-3 x 500 cm3 / 1000 cm3 = 0.10 mol
Number of moles of Cl- = 0.20 mol dm-3 x 500 cm3 / 1000 cm3 = 0.10 mol
For the 500 cm3 of 0.20 mol dm-3 Na2SO4 solution:
Number of moles of Na+ = 0.20 mol dm-3 x 500 cm3 / 1000 cm3 = 0.10 mol
Number of moles of SO42- = 0.20 mol dm-3 x 500 cm3 / 1000 cm3 = 0.10 mol
Now let's calculate the total number of moles for each species by summing up the moles from all the solutions:
Total number of moles of H+ = 0.10 mol + 0 mol (no H+ in NaCl and Na2SO4 solutions) = 0.10 mol
Total number of moles of Na+ = 0.10 mol + 0.10 mol = 0.20 mol
Total number of moles of Cl- = 0.10 mol + 0.10 mol = 0.20 mol
Total number of moles of SO42- = 0 mol (no SO42- in HCl solution) + 0.10 mol = 0.10 mol
Finally, we can calculate the final concentrations of each species in mol/L:
Final concentration of H+ = total number of moles of H+ / total volume of the mixed solution
= 0.10 mol / (1000 cm3 + 500 cm3 + 500 cm3) / 1000 cm3/dm3
= 0.10 mol / 2 L
= 0.05 mol/L
Final concentration of Na+ = total number of moles of Na+ / total volume of the mixed solution
= 0.20 mol / 2 L
= 0.10 mol/L
Final concentration of Cl- = total number of moles of Cl- / total volume of the mixed solution
= 0.20 mol / 2 L
= 0.10 mol/L
Final concentration of SO42- = total number of moles of SO42- / total volume of the mixed solution
= 0.10 mol / 2 L
= 0.05 mol/L
Therefore, the final concentrations in mol/L of H+, Na+, Cl-, and SO42- when the three solutions are mixed together are:
[H+] = 0.05 mol/L
[Na+] = 0.10 mol/L
[Cl-] = 0.10 mol/L
[SO42-] = 0.05 mol/L