Question

In the laboratory a student combines 44.9 mL of a 0.159 M barium nitrate solution with 16.2 mL of a 0.595 M barium chloride solution. What is the final concentration of barium cation?

Answer 1

[Ba2+] = 0.2746 M

Step-by-step explanation:

We have a 44.9 mL barium nitrate solution of 0.159M and a 16.2mL barium chloride solution of 0.595M

To find the concentration of the barium cation we use the following equation:

Concentration = moles of the solute / volumen of the solution

[Ba2+] = (44.9 * 10^-3 L * 0.159 M + 16.2 x 10^-3 L * 0.595 M) / ((44.9 mL +16.2mL) *10^-3 L)

After calculating this, we find that the concentration of the barium cation:

[Ba2+] = 0.2746 M