in what ratio 0.2M NaCl and 0.1M CaCl2 solutions are to be mixed so that in the resulting solution, the concentration of negative ions is 50% greater than the concentration of positive ions?

To determine the ratio in which the 0.2M NaCl and 0.1M CaCl2 solutions should be mixed, we need to consider the concentration of negative and positive ions.

Let's start by calculating the concentration of negative ions in the NaCl and CaCl2 solutions:

Concentration of negative ions in NaCl solution:
NaCl dissociates into Na+ and Cl- ions.
Since NaCl is a 1:1 electrolyte, the concentration of Cl- ions is equal to the concentration of Na+ ions.
Concentration of Cl- ions in 0.2M NaCl solution = 0.2M

Concentration of negative ions in CaCl2 solution:
CaCl2 dissociates into Ca2+ and 2Cl- ions.
Since CaCl2 is a 1:2 electrolyte, the concentration of Cl- ions is twice the concentration of Ca2+ ions.
Concentration of Cl- ions in 0.1M CaCl2 solution = 2 * 0.1M = 0.2M

Now, since we want the concentration of negative ions to be 50% greater than the concentration of positive ions, we need to find the concentration of positive ions.

Concentration of positive ions:
In the NaCl solution:
Concentration of Na+ ions = Concentration of Cl- ions = 0.2M

In the CaCl2 solution:
Concentration of Ca2+ ions = 0.1M

To satisfy the condition that the concentration of negative ions is 50% greater than the concentration of positive ions, we can set up the following equation:

0.2M (Concentration of Cl- ions) = 1.5 * 0.1M (Concentration of Ca2+ ions)

Now, let's solve for the concentrations:

0.2M (Concentration of Cl- ions) = 1.5 * 0.1M (Concentration of Ca2+ ions)
0.2M = 0.15M
Concentration of Cl- ions = 0.15M

Therefore, the ratio in which the solutions should be mixed is 0.15M NaCl solution to 0.1M CaCl2 solution.

To solve this problem, we need to consider the concentration of negative (Cl-) and positive (Na+ and Ca2+) ions separately and find their respective ratios in the resulting solution.

Let's assume we need to mix x mL of the 0.2M NaCl solution and y mL of the 0.1M CaCl2 solution. The total volume of the resulting solution would be (x + y) mL.

Now, let's calculate the concentration of negative ions in the resulting solution. In 0.2M NaCl, the concentration of Cl- ions is 0.2M. In 0.1M CaCl2, the concentration of Cl- ions is 0.1M * 2 (since there are two Cl- ions per CaCl2 molecule), which equals 0.2M as well.

Since we want the concentration of negative ions in the resulting solution to be 50% greater than the concentration of positive ions, we can set up the following equation:

0.2M = (0.5 * (x + y))M

Now, let's solve for x and y in terms of their ratio.

0.2 = 0.5x + 0.5y
0.2 = 0.5(x + y)
0.4 = x + y

We now have two equations:
x + y = 0.4 -----(1)
x/y = 0.2/0.1 = 2 -----(2)

To solve for x and y, we can use substitution or elimination method.

Let's use the elimination method by multiplying equation (2) by 2:
2x + 2y = 0.4 -----(3)

Subtract equation (1) from equation (3):
2x + 2y - (x + y) = 0.4 - 0.4
x + y = 0

This implies that x = -y

Now we can substitute this value of x in equation (1):
(-y) + y = 0.4
0 = 0.4

Since the equation is not satisfied, there is no valid solution for x and y that would give us a resulting solution with the desired concentration of ions.

Therefore, mixing the 0.2M NaCl and 0.1M CaCl2 solutions in the given ratio does not result in a solution where the concentration of negative ions is 50% greater than the concentration of positive ions.

I think what you want to do is this.

Let's take a volume (choose anything but I'll pick 100 mL of the 0.2M NaCl) and calculate the amount of CaCl2 needed to meet the problem.
100 mL of 0.2M NaCl + xmL of 0.1M CaCl2, we have
millimols + charge = 100*0.2 = 20 for Na in NaCl and 0.1x for Ca in CaCl2 for total of 20+0.1x and you want to multiply this by 1.5 so it will be 50% higher than the negative charge.

mmols negative charge is 100*0.2 = 20 for Cl in NaCl and 0.2x for the Cl in CaCl2 for total of 20+0.2x. Set these equal for the following:
1.5(20+0.1x) = 20+0.2x and solve for x = mL of 0.1M CaCl2 solution. You can go through this and you should, if for no other reason than to confirm my numbers. I obtained 200 mL of the CaCl2 so the ratio is 100 mL NaCl/200 mL CaCl2 or a ratio of 1 NaCl solution/2 CaCl2 solution. If I were you I would then plug these volumes back into the original solution and confirm that the negative charge = 50% greater than the positive charge. You may even want to try difference numbers, such as 500 mL and 1000 mL (still in the ratio of 1:2) and see if the equation holds.