A 7.00 g sample of KCl is dissolved in 81.0 mL of water. The resulting solution is then added to 28.0 mL of a 0.790 M CaCl2(aq) solution. Assuming that the volumes are additive, calculate the concentrations of each ion present in the final solution.
mols KCl = grams/molar mass = ?
mols CaCl2 = M x L = ?
mols K^+ = mols KCl
mols Ca^2+ = mols CaCl2
mols Cl^- = mols KCl + 2*mols CaCl2
Then (K^+) = mols K/L
(Ca^2+) = mols Ca/L
(Cl^-) = mols Cl/L
Note: total volume is 81 mL + 28 mL = ? mL and convert to L.
To calculate the concentrations of each ion present in the final solution, we need to find the number of moles of each ion.
First, let's calculate the number of moles of KCl:
Molar mass of KCl = (39.10 g/mol of K) + (35.45 g/mol of Cl) = 74.55 g/mol
Number of moles of KCl = 7.00 g / 74.55 g/mol ≈ 0.094 mol
Next, let's calculate the number of moles of CaCl2:
Number of moles of CaCl2 = concentration of CaCl2 × volume of CaCl2 solution
= 0.790 mol/L × 0.0280 L ≈ 0.022 mol
Now, let's calculate the total volume of both solutions:
Total volume of solution = volume of KCl solution + volume of CaCl2 solution
= 0.0810 L + 0.0280 L = 0.109 L
Now, let's calculate the concentration of K+ ions in the final solution:
Concentration of K+ ions = (moles of K+ ions) / (total volume of solution)
= (0.094 mol) / (0.109 L) ≈ 0.862 M
Similarly, let's calculate the concentration of Cl- ions in the final solution:
Concentration of Cl- ions = (moles of Cl- ions) / (total volume of solution)
= (2 × moles of CaCl2) / (total volume of solution)
= (2 × 0.022 mol) / (0.109 L) ≈ 0.403 M
Therefore, the concentrations of K+ and Cl- ions in the final solution are approximately 0.862 M and 0.403 M, respectively.
To calculate the concentrations of each ion present in the final solution, we need to first determine the number of moles of each compound and then divide by the total volume of the solution.
Let's begin by calculating the number of moles of KCl:
1. Use the formula: Moles = mass / molar mass
The molar mass of KCl is 74.55 g/mol.
Moles of KCl = 7.00 g / 74.55 g/mol = 0.094 mol
Next, let's calculate the number of moles of CaCl2:
1. Use the formula: Moles = concentration x volume
The concentration of CaCl2 is given as 0.790 M, and the volume of the CaCl2 solution is 28.0 mL.
Moles of CaCl2 = 0.790 mol/L x 0.0280 L = 0.022 mol
Now we can determine the total volume of the final solution:
1. Use the formula: Total volume = volume of KCl solution + volume of CaCl2 solution
The volume of the KCl solution is given as 81.0 mL, and the volume of the CaCl2 solution is 28.0 mL.
Total volume = 81.0 mL + 28.0 mL = 109.0 mL = 0.109 L
Finally, we can calculate the concentrations of each ion:
Concentration of K+: Moles of K+ / Total volume
Since KCl dissociates into one K+ ion, the concentration is equal to the number of moles of KCl.
Concentration of K+ = 0.094 mol / 0.109 L = 0.862 M
Concentration of Cl-: Moles of Cl- / Total volume
Since KCl dissociates into one Cl- ion, the concentration is also equal to the number of moles of KCl.
Concentration of Cl- = 0.094 mol / 0.109 L = 0.862 M
Concentration of Ca2+: Moles of Ca2+ / Total volume
Since CaCl2 dissociates into two Ca2+ ions, we need to multiply the number of moles of CaCl2 by 2.
Moles of Ca2+ = 2 x 0.022 mol = 0.044 mol
Concentration of Ca2+ = 0.044 mol / 0.109 L = 0.404 M
Therefore, the concentrations of each ion present in the final solution are as follows:
- K+: 0.862 M
- Cl-: 0.862 M
- Ca2+: 0.404 M