In the laboratory, a student combines 28.3 ml of a 0.489 m manganese(ii) acetate solution with 12.5 ml of a 0.339 m manganese(ii) sulfate solution. what is the final concentration of manganese(ii) cation?

M = 0.441 M

Explanation:

In this case, we have two solutions that involves the Manganese II cation;

We have Mn(CH₃COOH)₂ and MnSO₄

In both cases, the moles of Mn are the same in reaction as we can see here:

Mn(CH₃COO)₂ <> Mn²⁺ + 2CH₃COO⁻

MnSO₄ <> Mn²⁺ + SO₄²⁻

Therefore, all we have to do is calculate the moles of Mn in both solutions, do the sum and then, calculate the concentration with the new volume:

moles of MnAce = 0.489 * 0.0283 = 0.0138 moles

moles MnSulf = 0.339 * 0.0125 = 0.0042 moles

the total moles are:

moles of Mn²⁺ = 0.0138 + 0.0042 = 0.018 moles

Finally the concentration: 12.5 + 28.3 = 40.8 mL or 0.0408 L

M = 0.018 / 0.0408

M = 0.441 M

This would be the final concentration of the manganese after the mixing of the two solutions