In the laboratory a student combines 25.9 mL of a 0.223 M magnesium bromide solution with 23.3 mL of a 0.615 M magnesium acetate solution.
What is the final concentration of magnesium cation ?
..... M
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.
To find the final concentration of the magnesium cation, we need to calculate the moles of magnesium cation in each solution separately and then add them together. We can then divide the total moles by the total volume of the combined solutions to get the final concentration.
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.