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
To calculate the final concentrations of H+, Na+, Cl-, and SO42- when the three solutions are mixed together, we need to use the principles of stoichiometry and the concept of solution dilutions.
First, let's convert the volumes of the solutions to liters:
1000 cm3 = 1000/1000 = 1 L
500 cm3 = 500/1000 = 0.5 L
Now, we can start by calculating the number of moles of each ion in each solution.
1. HCl solution:
Volume = 1 L
Concentration = 0.10 mol/L
Number of moles of H+ and Cl- in the HCl solution = Volume * Concentration = 1 L * 0.10 mol/L = 0.10 mol
2. NaCl solution:
Volume = 0.5 L
Concentration = 0.20 mol/L
Number of moles of Na+ and Cl- in the NaCl solution = Volume * Concentration = 0.5 L * 0.20 mol/L = 0.10 mol
3. Na2SO4 solution:
Volume = 0.5 L
Concentration = 0.20 mol/L
Number of moles of Na+ in the Na2SO4 solution = 2 * Volume * Concentration = 2 * 0.5 L * 0.20 mol/L = 0.20 mol
Number of moles of SO42- in the Na2SO4 solution = Volume * Concentration = 0.5 L * 0.20 mol/L = 0.10 mol
Next, we need to determine the total moles of each ion after mixing the solutions together. Since HCl and NaCl both dissociate completely in water, the total moles of H+ and Cl- ions will be the sum of the moles from the HCl and NaCl solutions.
Total moles of H+ and Cl- ions = Moles of H+ and Cl- from HCl solution + Moles of H+ and Cl- from NaCl solution
= 0.10 mol + 0.10 mol = 0.20 mol
For Na+ and SO42-, we need to consider the stoichiometry of Na2SO4. Since Na2SO4 dissociates into 2 moles of Na+ and 1 mole of SO42-, the total moles of Na+ and SO42- ions will be the sum of the moles from the NaCl solution and half the moles from the Na2SO4 solution.
Total moles of Na+ ions = Moles of Na+ from NaCl solution + (0.5 * Moles of Na+ from Na2SO4 solution)
= 0.10 mol + (0.5 * 0.20 mol) = 0.20 mol
Total moles of SO42- ions = 0.5 * Moles of SO42- from Na2SO4 solution
= 0.5 * 0.10 mol = 0.05 mol
Finally, we can calculate the final concentrations of each ion by dividing the total moles by the total volume of the mixed solution, which is the sum of the initial volumes of the three solutions.
Final concentration of H+ = Total moles of H+ ions / Total volume of mixed solution
= 0.20 mol / (1 L + 0.5 L + 0.5 L) = 0.20 mol / 2 L = 0.10 mol/L
Final concentration of Na+ = Total moles of Na+ ions / Total volume of mixed solution
= 0.20 mol / 2 L = 0.10 mol/L
Final concentration of Cl- = Total moles of Cl- ions / Total volume of mixed solution
= 0.20 mol / 2 L = 0.10 mol/L
Final concentration of SO42- = Total moles of SO42- ions / Total volume of mixed solution
= 0.05 mol / 2 L = 0.025 mol/L
Therefore, the final concentrations in mol/L of H+, Na+, Cl-, and SO42- in the mixed solution are:
H+: 0.10 mol/L
Na+: 0.10 mol/L
Cl-: 0.10 mol/L
SO42-: 0.025 mol/L