Respuesta :
Answer:
1 M
Explanation:
Magnesium chloride will furnish chloride ions as:
[tex]MgCl_2\rightarrow Mg^{2+}+2Cl^-[/tex]
Given :
Moles of magnesium chloride = 0.20 mol
Thus, moles of chlorine furnished by magnesium chloride is twice the moles of magnesium chloride as shown below:
[tex]Moles =2\times 0.20\ moles[/tex]
Moles of chloride ions by magnesium chloride = 0.40 moles
Potassium chloride will furnish chloride ions as:
[tex]KCl\rightarrow K^{+}+Cl^-[/tex]
Given :
Moles of potassium chloride = 0.10 moles
Thus, moles of chlorine furnished by potassium chloride is same as the moles of potassium chloride as shown below:
Moles of chloride ions by potassium chloride = 0.10 moles
Total moles = 0.40 + 0.10 moles = 0.50 moles
Given, Volume = 500 mL = 0.5 L (1 mL = 10⁻³ L)
Concentration of chloride ions is:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity_{Cl^-}=\frac{0.50}{0.5}[/tex]
The final concentration of chloride anion = 1 M
The concentration of the chloride ion, Cl¯ in the solution is 1 M
To obtain the answer to the question given above, we'll begin by calculating the number of mole of chloride ion, Cl¯ produced by each compound in the solution. This can be obtained as follow:
For MgCl₂:
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
From the balanced equation above,
1 mole of MgCl₂ produced 2 moles of Cl¯.
Therefore, 0.2 mole of MgCl₂ will produce = 0.2 × 2 = 0.4 mole of Cl¯
For KCl:
KCl(aq) —> K⁺(aq) + Cl¯(aq)
From the balanced equation above,
1 mole of KCl produced 1 mole of Cl¯
Therefore,
0.1 mole of KCl will also produce 0.1 mole of Cl¯.
Next, we shall determine the total mole of Cl¯ in the solution.
Mole of Cl¯ from MgCl₂ = 0.4 mole
Mole of Cl¯ from KCl = 0.1 mole
Total mole of Cl¯ =?
Total mole of Cl¯ = 0.4 + 0.1
Total mole of Cl¯ = 0.5 mole
Finally, we shall determine the concentration of Cl¯ in the solution.
Total mole of Cl¯ = 0.5 mole
Volume = 500 mL = 500 / 1000 = 0.5 L
Concentration of Cl¯ =?
Concentration = mole / Volume
Concentration of Cl¯ = 0.5 / 0.5
Concentration of Cl¯ = 1 M
Therefore, the concentration of Cl¯ in the solution is 1 M
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