Question

A solution is prepared by dissolving 0.24 mol of butanoic acid and 0.25 mol of sodium butanoate in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly. The pH does not increase drastically because the NaOH reacts with the ________ present in the buffer solution. The Ka of butanoic acid is 1.5 × 10-5.

Answer 1

Answer:

Butanoic acid present in solution

Explanation:

In this case, we have a buffer solution of butanoic acid and sodium butanoate.  In other words a reaction like this:

HC₄H₇O₂ + H₂O <------> C₄H₇O₂⁻ + H₃O⁺   Ka = 1.5x10⁻⁵

The low value of Ka means that this is a weak acid. So, after this, the NaOH is added to the solution.

The NaOH is a really strong base, so we might expect that the pH of the solution increase drastically, however this do not occur.

The reason for this is because the first thing to happen in this reaction is an acid base reaction.

The NaOH react with the butanoic acid still present in solution, because is a weak acid, so in solution, this acid is not completely dissociated into it's respective ions. So the butanoic acid reacts with the NaOH and the products:

HC₄H₇O₂ + NaOH <------> Na⁺C₄H₇O₂⁻ + H₂O

So, because of this, the pH increase but not much.