Asolution is prepared by adding 27.00 ml of 2.19 m mgcl2 to enough water to make 122.00 ml. what is the cl- concentration of 40.00 ml of the dilute solution?

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3.88x10⁻²

Explanation:

First, the amount of moles initially in the volume of the concentrated MgCl2 solution is determined:

1000 mL solution 2.19 moles of MgCl₂

27.00 mL solution X = 5.91x10⁻² mol of MgCl₂

Calculation: 27.00 mL x 2.19 moles / 1000 mL = 0.05913 moles ≡ 5.91x10⁻² mol of MgCl₂

This is the amount of moles of MgCl₂ in both the initial and the diluted solution, since only water was added to the first solution, therefore, in 122.00 mL of the new solution, there are 5.91x10⁻² mol of MgCl₂. Now it is necessary to calculate the amount of moles in 40 mL of the new solution prepared:

122.00 mL of solution 5.91x10⁻² mol of MgCl₂

40.00 mL of solution X = 1.94x10⁻² of MgCl₂

Calculation: 40.00 mL x 5.91x10⁻² mol / 122.00 mL = 0.01939 moles ≡ 1.94x10⁻² MgCl₂

Now, when the MgCl₂ compound is in an aqueous solution, it dissociates into the ions that form it as follows:

MgCl₂ ⇒ Mg⁺² (aq) + 2 Cl⁻ (aq)

Therefore, for each mole of MgCl₂ that dissolves in water, it will dissociate producing two moles of Cl⁻ chloride ion. Mathematically:

1 mole of MgCl₂ 2 moles of Cl⁻

1.94x10⁻² moles of MgCl₂ X = 3.88x10⁻² moles of Cl⁻

Calculation: 1.94x10⁻² moles x 2 moles / 1 mole = 0.03877 ≡ 3.88x10⁻² moles of Cl⁻

Therefore, in 40.00 mL of the new solution, there will be a concentration of 3.88x10⁻² moles of Cl⁻

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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