Indicate the concentration of each ion present in the solution formed by mixing the following. (Assume that the volumes are additive.)
(a) 50 mL of 0.100 M HCl and 10.0 mL of 0.510 M HCl
(b) 15.0 mL of 0.309 M Na2SO4 and 38.6 mL of 0.200 M KCl
(c) 3.50 g of NaCl in 40.2 mL of 0.372 M CaCl2 solution
(A) moles HCl1 = M x L
moles HCl2 = M x L
moles mixture = total moles/total L.
B and C are about the same but the salts are not 1:1 therefore, you determine moles Na2SO4, for example, and multiply by 1 for moles Na^+ and multiply by 1 for sulfate ion.
(a) To find the concentration of each ion present in the solution formed by mixing the two solutions of HCl, we can use the principles of stoichiometry and the fact that volumes are additive.
For the first solution, 50 mL of 0.100 M HCl, we know that the concentration of H+ ions is 0.100 M. Since HCl is a strong acid, it dissociates completely in water to form H+ and Cl- ions.
For the second solution, 10.0 mL of 0.510 M HCl, we again have a concentration of 0.510 M HCl. So, in this case as well, the concentration of H+ ions is 0.510 M.
When we mix the two solutions, the volumes sum up. So the total volume of the combined solution is 50 mL + 10.0 mL = 60.0 mL.
The total concentration of H+ ions in the combined solution is the sum of the two individual concentrations:
0.100 M + 0.510 M = 0.610 M.
Since HCl is a strong acid that completely dissociates in water, the concentration of Cl- ions will be the same as the concentration of H+ ions:
0.610 M.
Therefore, the concentration of each ion present in the solution formed by mixing 50 mL of 0.100 M HCl and 10.0 mL of 0.510 M HCl is:
H+ ions: 0.610 M
Cl- ions: 0.610 M.
(b) Applying the same principles, we can determine the concentration of each ion present in the solution formed by mixing 15.0 mL of 0.309 M Na2SO4 and 38.6 mL of 0.200 M KCl.
For the first solution, 15.0 mL of 0.309 M Na2SO4, we know that the concentration of Na+ ions is 2 * 0.309 M = 0.618 M (because there are two Na+ ions per formula unit of Na2SO4) and the concentration of SO4^2- ions is 0.309 M.
For the second solution, 38.6 mL of 0.200 M KCl, we can directly extract the concentrations of K+ ions (0.200 M) and Cl- ions (0.200 M).
When we mix the two solutions, the volumes add up. So the total volume of the combined solution is 15.0 mL + 38.6 mL = 53.6 mL.
The total concentration of Na+ ions in the combined solution is the sum of the two individual concentrations:
0.618 M.
The total concentration of K+ ions in the combined solution is the same as the concentration of KCl:
0.200 M.
For the anions, since Na2SO4 dissociates to give two SO4^2- ions per formula unit, the total concentration of SO4^2- ions in the combined solution is twice the concentration of Na2SO4:
2 * 0.309 M = 0.618 M.
The total concentration of Cl- ions in the combined solution is the sum of the two individual concentrations:
0.200 M.
Therefore, the concentration of each ion present in the solution formed by mixing 15.0 mL of 0.309 M Na2SO4 and 38.6 mL of 0.200 M KCl is:
Na+ ions: 0.618 M
K+ ions: 0.200 M
SO4^2- ions: 0.618 M
Cl- ions: 0.200 M.
(c) To determine the concentration of each ion present in the solution formed by mixing 3.50 g of NaCl in 40.2 mL of 0.372 M CaCl2 solution, we need to calculate the amount (in moles) of NaCl and then divide by the final volume.
The molar mass of NaCl is 58.44 g/mol, so 3.50 g of NaCl is equal to 3.50 g / 58.44 g/mol = 0.05988 mol.
The final volume of the combined solution is 40.2 mL.
The concentration of Na+ ions in the solution formed by dissolving 3.50 g of NaCl in 40.2 mL of 0.372 M CaCl2 is:
0.05988 mol / 0.0402 L = 1.487 M.
The concentration of Cl- ions in the solution formed by dissolving 3.50 g of NaCl in 40.2 mL of 0.372 M CaCl2 is the same as the concentration of Na+ ions:
1.487 M.
Therefore, the concentration of each ion present in the solution formed by mixing 3.50 g of NaCl in 40.2 mL of 0.372 M CaCl2 solution is:
Na+ ions: 1.487 M
Cl- ions: 1.487 M.
To find the concentration of each ion in a solution formed by mixing different solutions, you can use the principles of dilution and the concept of moles and stoichiometry of the compounds involved.
(a) For the first mixture:
1. Determine the total volume of the solution formed by adding the two initial volumes: 50 mL + 10 mL = 60 mL.
2. Convert the initial volumes to liters: 50 mL = 0.050 L and 10 mL = 0.010 L.
3. Calculate the moles of HCl in each solution using the formula: moles = concentration × volume.
For the first solution: moles = 0.100 M × 0.050 L = 0.005 mol HCl
For the second solution: moles = 0.510 M × 0.010 L = 0.0051 mol HCl
4. Add the moles of HCl from both solutions to find the total moles: 0.005 mol + 0.0051 mol = 0.0101 mol HCl.
5. Use the total volume of the solution to calculate the new concentration of HCl: concentration = moles / volume.
concentration = 0.0101 mol / 0.060 L = 0.1683 M HCl.
Since HCl dissociates into H+ and Cl- ions in solution in a 1:1 ratio, the concentration of H+ and Cl- ions will be the same as the concentration of HCl: 0.1683 M.
(b) For the second mixture:
1. Determine the total volume of the solution formed by adding the two initial volumes: 15 mL + 38.6 mL = 53.6 mL.
2. Convert the initial volumes to liters: 15 mL = 0.015 L and 38.6 mL = 0.0386 L.
3. Calculate the moles of Na2SO4 and KCl in each solution using the formula: moles = concentration × volume.
For Na2SO4:
Moles = 0.309 M × 0.015 L = 0.004635 mol Na2SO4
For KCl:
Moles = 0.200 M × 0.0386 L = 0.00772 mol KCl
4. Use stoichiometry to determine the moles of Na+ and SO4 2- ions present, based on the chemical equation:
Na2SO4(aq) → 2 Na+(aq) + SO4 2-(aq)
Moles of Na+ ions = 2 × 0.004635 mol = 0.00927 mol Na+
Moles of SO4 2- ions = 0.004635 mol SO4 2-
5. Similarly, determine the moles of K+ and Cl- ions:
Moles of K+ ions = 0.00772 mol K+
Moles of Cl- ions = 0.00772 mol Cl-
6. Calculate the new concentration of each ion: concentration = moles / volume.
For Na+ ions: concentration = 0.00927 mol / 0.0536 L ≈ 0.173 M
For SO4 2- ions: concentration = 0.004635 mol / 0.0536 L ≈ 0.086 M
For K+ ions: concentration = 0.00772 mol / 0.0536 L ≈ 0.144 M
For Cl- ions: concentration = 0.00772 mol / 0.0536 L ≈ 0.144 M
(c) For the third mixture:
1. Calculate the moles of NaCl in the given amount (3.50 g) using its molar mass.
Moles = mass / molar mass
Moles of NaCl = 3.50 g / (22.99 g/mol + 35.45 g/mol) ≈ 0.066 mol NaCl
2. Calculate the total volume of the solution formed by adding the initial volume (40.2 mL) to the obtained volume of the NaCl solution.
Total volume = 40.2 mL + (0.066 mol NaCl × (1 L/0.372 mol CaCl2)) ≈ 41.2 mL
3. Convert the total volume to liters: 41.2 mL = 0.0412 L.
4. Calculate the concentration of CaCl2 using the formula: concentration = moles / volume.
Concentration = 0.372 M (given)
5. Since NaCl dissociates into Na+ and Cl- ions in solution in a 1:1 ratio, the concentration of Na+ and Cl- ions will be the same:
Concentration of Na+ ions = 0.066 M
Concentration of Cl- ions = 0.066 M
I hope this helps! Let me know if you have any further questions.