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.
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 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.
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