JSmol JSmol JSmol JSmol Added 2.0 M Acetic Acid Added 2.0 M CH3COONa CH3COOH (aq
ID: 712740 • Letter: J
Question
JSmol JSmol JSmol JSmol Added 2.0 M Acetic Acid Added 2.0 M CH3COONa CH3COOH (aq) H20 () H3o+ (aq) CH3COO (aq) We have seen that adding the common ion (salt of the conjugate base) to a weak acid equilibrium changes the pH. Imagine a solution in which you have created a solution that is 2.0 M in acetic acid and 2.0 M in sodium acetate (the purple atom is the sodium cation). In this case the concentration of the weak acid and it's conjugate base are exactly equal. In truth, a small amount of acetate anion (conjugate base) is also coming from the dissociation of the weak acid, but this is negligible compared to the added sodium acetate Check all of the statements that will be true if these two concentrations are equal. Hint: write out the Ka expression to help you think it through. The Na+ cation does not appear in the equilibrium because it is a spectator ion and is only the counter ion for the acetate anion All of the acetic acid will become deprotonated by water so the final solution contains ony aqueous h aron un ions and acetate ons. The pKa for the acid = the pH of this solution. Since acetic acid is a weak acid, acetate anion is considered to be a strong base The [H30+-Ka in this solution. The pH of this solution will be lower than that of a 2.0 M solution of acetic acid.Explanation / Answer
The acid ionization equation is given as
CH3COOH (aq) + H2O (l) <=======> H3O+ (aq) + CH3COO- (aq)
The ionization constant Ka is expressed as
Ka = [H3O+][CH3COO-]/[CH3COOH] (concentration of liquid water is not included in the expression for the equilibrium constant).
We know that CH3COOH is a weak acid and hence, CH3COO- is the conjugate base of the weak acid. The acid-base equilibrium involves only CH3COOH/CH3COO-. Na+ balances the charge on CH3COO- and is therefore, a counter ion. Na+ doesn’t appear in the equilibrium constant expression and hence, the first statement is true.
The acid-base system is at equilibrium and hence, the system will contain both CH3COOH and CH3COO- at equilibrium. Therefore, the second statement is false.
We have [CH3COOH] = [CH3COO-] = 2.0 M; therefore,
Ka = [H3O+]; consequently,
-log (Ka) = -log [H3O+]
=====> pKa = pH
Therefore, the third and fifth statements are true.
Acetic acid is a weak acid and the conjugate base is a weak base; hence, the fourth statement is false.
When [CH3COOH] = [CH3COO-] = 2.0 M, we have pH = pKa. The pKa of acetic acid is about 4.75; hence, the pH 4.75 and thus, the pH is higher than than of 2.0 M acetic acid. Therefore, the last statement is false.
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