A student mixes 0.50 L of 0.0020 M Ba (NO3) 2 solution with 0.50 L of 0.0080 M Na2SO4 solution. Which equation shows the proper set-up to calculate molar concentration of Ba2 + ions in the new solution?

(0.0020 M) (1 L) = (? M)

(0.0020 M) (0.5 L) = (? M) (1 L)

(0.0020 M) (0.5 L) = (0.0080 M) (1 L)

(0.0020 M) (0.5 L) = (0.0080 M) (0.5) (? L)

Question

Juliette

Chemistry

11 May, 16:40

A student mixes 0.50 L of 0.0020 M Ba (NO3) 2 solution with 0.50 L of 0.0080 M Na2SO4 solution. Which equation shows the proper set-up to calculate molar concentration of Ba2 + ions in the new solution?

(0.0020 M) (1 L) = (? M)

(0.0020 M) (0.5 L) = (? M) (1 L)

(0.0020 M) (0.5 L) = (0.0080 M) (1 L)

(0.0020 M) (0.5 L) = (0.0080 M) (0.5) (? L)

+3

Answers (1)

Know the Answer?

Jaylee Bruce · Answer 1

(0.0020 M) * (0.5 L) = (? M) * (1 L)

Explanation:

To find the concentration of the final solution when you concentrate or dilute a particular solution is done by using the following formula:

initial concentration * initial volume = final concentration * final volume

The question is asking about the concentration of barium ions in the final solution.

initial concentration of Ba²⁺ = 0.0020 M

initial volume = 0.50 L

final concentration = unknown (?)

final volume = volume of the initial solution + volume of the second solution

final volume = 0.50 + 0.50 = 1 L

You insert the values in the equation and you find the answer asked by the question.