mols Mg from MgBr2 = M x L = ?
mols Mg from Mg(Ac)2 = M x L
(Mg) in final solution = total mols Mg/total L solution.
What is the final concentration of magnesium cation ?
..... M
mols Mg from Mg(Ac)2 = M x L
(Mg) in final solution = total mols Mg/total L solution.
First, let's calculate the moles of magnesium cation in the magnesium bromide solution:
Moles of magnesium cation in magnesium bromide solution = concentration (M) x volume (L)
= 0.223 M x (25.9 mL / 1000 mL/L) (converting mL to L)
= 0.0057807 moles
Next, let's calculate the moles of magnesium cation in the magnesium acetate solution:
Moles of magnesium cation in magnesium acetate solution = concentration (M) x volume (L)
= 0.615 M x (23.3 mL / 1000 mL/L) (converting mL to L)
= 0.0142995 moles
Now, we add the moles from both solutions together:
Total moles of magnesium cation = 0.0057807 moles + 0.0142995 moles
= 0.0200802 moles
Finally, we divide the total moles by the total volume of the combined solutions to get the final concentration:
Final concentration of magnesium cation = Total moles / Total volume (L)
= 0.0200802 moles / ((25.9 mL + 23.3 mL) / 1000 mL/L) (converting mL to L)
= 0.0200802 moles / 0.0492 L
= 0.407 M
Therefore, the final concentration of the magnesium cation is 0.407 M.