The common-ion effect is defined as the shift in an ionic equilibrium produced b

Question

The common-ion effect is defined as the shift in an ionic equilibrium produced by the addition of a solute that provides an ion common to the equilibrium. Compare the following two solutions:

Solution 1: 0.100 M C2H4O2
Solution 2: 5mL 0.100 M C2H4O2 + 5mL 0.100 M HCl

The [H3O+] ions from the strong acid in solution 2 force the equilibrium C2H4O2 ? H2O + C2H3O2- almost completely to the left. As a result, the pH is determined by the H3O+ ions from the strong acid. Calculate the pH of Solution 1 (ka = 1.738 x 10-5) and 2 (assume HCl dominate for calculation)

What is the pH level for Solution 1 and Solution 2?

Explanation / Answer

Part 1:

The equilibrium constant expression will be: Ka = [C2H3O2-][H3O+]/[C2H4O2]

Replacing the values that we have in the table:

Ka=x2/0,100-x

Assuming that the x value is smalest than 0.100, we can make the math in the following way:

x2= (1,738x10-5)(0.100) = 1,738x10-5

Now we need to calculate the square root of this number to obtain x:

x = 0,001318

This is the value for the H3O+ concentration.

The pH is -log([H3O+ ]), so the pH for the Solution 1 is: 2,88.

Part 2

In the problem mention that we can assume that the HCl dominate the solution, so the pH is define for the H3O+ concentration in HCl solution.

pH= - log(0.100) = 1,00

C2H4O2 + H2O = C2H3O2- + H3O+ Initial Conditions 0,100 Change -x x x Equilibrium 0,100-x x x

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