Question

A solution is prepared by dissolving 0.23 mol of lactic acid and 0.27 mol of sodium lactate 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 lactic acid is 1.4 ⋅ 10-4.

Answer 1

Answer:

The pH does not increase drastically because the NaOH reacts with the lactic acid present in the buffer solution.

Explanation:

A buffer is the equilibrium that is produced by yje misture of a weak acid with its conjugate base or vice versa.

In the problem, the equilibrium is:

Lactic acid ⇄ Sodium lactate + H⁺

The addition of a strong base as NaOH produce the reaction with lactate acid to form sodium lactate and water, thus:

Lactic acid + NaOH ⇄ Sodium lactate + H₂O

The pH does not increase drastically because, as you can see, there is no formation of free OH⁻ that are responsible of pH increasing

Answer 2

Explanation:

The solution of the lactic acd and sodium lactate is referred to as a buffer solution.

A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. In this case, the weak acid is the lactic acid and the conjugate base is the sodium lactate.

Buffer solutions are generally known to resist change in pH values.

When a strong base (in this case, NaOH) is added to the buffer, the lactic acid will give up its H+ in order to transform the base (OH-) into water (H2O) and the conjugate base, so we have:

HA + OH- → A- + H2O.

Since the added OH- is consumed by this reaction, the pH will change only slightly.

The NaOH reacts with the weak acid present in the buffer sollution.