Write the balanced dissociation equation for MgCl2. (1 mark)
MgCl2(s) + H2O(l) ==> Mg^+2(aq) + 2Cl^-(aq)
b) In 300.0 mL of a 0.50 M MgCl2 solution, what would be the concentration of the chloride ions?
In 300.0 mL of a 0.50 M MgCl2 solution, what would be the concentration of the chloride ions?
(MgCl2) is 0.50 M.
There are two Cl^- for each MgCl2; therefore, the Cl will be twice the MgCl2.
The equation for part a is correct.
To find the concentration of chloride ions in the MgCl2 solution, we need to use the balanced dissociation equation for MgCl2:
MgCl2(s) + H2O(l) → Mg^2+(aq) + 2Cl^-(aq)
From the equation, we can see that 1 molecule of MgCl2 dissociates into 2 chloride ions (Cl^-). So the concentration of chloride ions is twice the concentration of MgCl2.
Given that the volume of the solution is 300.0 mL and the concentration of MgCl2 is 0.50 M, we can calculate the concentration of chloride ions.
Concentration of chloride ions = 2 * concentration of MgCl2
Concentration of chloride ions = 2 * 0.50 M
Concentration of chloride ions = 1.00 M
Therefore, the concentration of chloride ions in the solution is 1.00 M.
To find the concentration of chloride ions in the MgCl2 solution, we need to use the balanced dissociation equation for MgCl2:
MgCl2(s) + H2O(l) → Mg^2+(aq) + 2Cl^-(aq)
From the equation, we see that for every one mole of MgCl2 that dissociates, we obtain two moles of chloride ions, Cl^-. Therefore, the concentration of the chloride ions is twice the concentration of MgCl2.
Given that the MgCl2 solution has a concentration of 0.50 M, we can calculate the concentration of chloride ions as follows:
Concentration of chloride ions = 2 × concentration of MgCl2
Concentration of chloride ions = 2 × 0.50 M
Concentration of chloride ions = 1.00 M
Therefore, the concentration of chloride ions in the 300.0 mL of a 0.50 M MgCl2 solution would be 1.00 M.