a solution is made by combining 5 mL of .2M HC2H3O2, 5 mL of .2M NaC2H3O2, and 10 ml of water. calculate the molarity of HC2H3O2 and of NaC2H3O2 in this new solution

You have diluted the HC2H3O2 solution and 5 mL of the NaC2H3O2 solution to 20 mL; therefore, the new concns are

0.2M x (5/20) = ??

To calculate the molarity of HC2H3O2 and NaC2H3O2 in the new solution, we need to use the concept of molarity and understand the principles of dilution.

Molarity (M) is defined as the number of moles of solute dissolved in one liter of solution. In this case, we need to determine the molarity of HC2H3O2 and NaC2H3O2 in the resulting solution.

First, let's calculate the amount of moles of HC2H3O2 and NaC2H3O2 in their respective original solutions. To do this, we'll use the formula:

moles = molarity × volume (in liters)

For HC2H3O2:
moles of HC2H3O2 = 0.2 M × 0.005 L
moles of HC2H3O2 = 0.001 mol

For NaC2H3O2:
moles of NaC2H3O2 = 0.2 M × 0.005 L
moles of NaC2H3O2 = 0.001 mol

Now, let's calculate the total volume of the resulting solution by summing the volumes of the individual components:

total volume = 5 mL + 5 mL + 10 mL = 20 mL = 0.02 L

Next, we need to calculate the moles of HC2H3O2 and NaC2H3O2 in the resulting solution.

moles of HC2H3O2 in resulting solution = 0.001 mol
moles of NaC2H3O2 in resulting solution = 0.001 mol

Since the total volume is given as 0.02 L, we can calculate the molarity of HC2H3O2 and NaC2H3O2 in the resulting solution using the formula:

molarity = moles / volume (in liters)

For HC2H3O2:
molarity of HC2H3O2 = 0.001 mol / 0.02 L
molarity of HC2H3O2 = 0.05 M

For NaC2H3O2:
molarity of NaC2H3O2 = 0.001 mol / 0.02 L
molarity of NaC2H3O2 = 0.05 M

Therefore, the molarity of both HC2H3O2 and NaC2H3O2 in the resulting solution is 0.05 M.