Answer:
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
