01:160:162 Lecture Notes - Lecture 5: Sodium Hydroxide, Equivalence Point, Titration

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Resists ph change by neutralizing added acid or base. Solution that resists ph change when small amounts of hydroxide or hydronium ions are added o. Significant amounts of both weak acid and conjugate base. Significant amounts of both a weak base and a conjugate acid. Acid component can neutralize small amounts of oh. Basic component can neutralize small amounts of h3o o. The buffer components (ha and a) are able to neutralize small amounts of added oh or h3o+ Ex. a buffer that has (h2co3 and hco3-: when additional base is added, weak acid reacts with the base, neutralizing it, when additional acid is added, conjugate base reacts with the acid, neutralizing. Buffer can maintain a nearly constant ph it o o. * a buffer should have roughly equal amounts of conjugate base / acid with weak acid or base. The ph of a buffer solution can be calculated with this equation.

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Experiment 18: Buffers Purpose In this experiment you will prepare several buffers and investigate their properties. Discussion A solution of a weak acid and one of its salts, in approximately equal concentrations, resists changes in pH when acid or base is added to it. (Note: The pH of a solution is a measure of the concentration of free H3Ot ion; it does not indicate concentration of combined acid such as HA.) The same is true for a solution of a weak base and one of its salts. A solution with this property is called a buffer (see illustration next page). Bufter systems are extremely important in many chemical and biological systems. For example, your blood contains a buffer system utilizing carbonic acid and bicarbonate ion. The important equilibrium in an acidic buffer is HA (aq) H20 ) A (a)0 () acid conjugate base in which the concentrations of acid and conjugate base are approximately equal. If a small amount of free acid (H30) is added to this solution, it reacts with the conjugate base A (reaction 1 in reverse). The equilibrium constant for this reaction is very large, so the reaction proceeds to near completion, absorbing the added acid. Similarly, added base (OH-) reacts with the acid HA as shown below: 2. HA (aq) + OH-(aq) ATaq) + H2O (1) This reaction also proceeds to completion, effectively absorbing any added base. Since the H30 and OH- concentrations are kept constant, the solution thus resists pH changes when an acid or base is added to it. A similar argument can be made for a basic buffer The actual pH of a buffer can be determined from the value of the Ka and the concentrations of the acid and conjugate base obtained from an initial-change- equilibrium table. However, a shortcut for buffers is available, and is shown below. We begin by writing and rearranging the equilibrium constant expression for the acid dissociation. If the -log of both sides of equation (3) is taken, a relationship involving pH is then obtained. H30IA] HA] 3. HA HAI -loglH30'1- -log Ka A- 141

Â± pH Changes in Buffers

When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. Buffers resist change in pH following the addition of acid or base. A buffer solution prepared from a weak acid (HA) and its conjugate base (Aâ) is represented as

HA(aq)âH+(aq)+Aâ(aq)

The buffer will follow Le ChÃ¢telier's principle. If acid is added, the reaction shifts to consume the added H+, forming more HA. When base is added, the base will react with H+, reducing its concentration. The reaction then shifts to replace H+ through the dissociation of HA into H+ and Aâ. In both instances, [H+] tends to remain constant.

The pH of a buffer is calculated by using the Henderson-Hasselbalch equation:

pH=pKa+log[Aâ][HA]

Part A

What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.609 mol of NaAin 2.00 L of solution? The dissociation constant Ka of HA is 5.66Ã10â7.

Express the pH numerically to three decimal places.

Hints

pH =

6.424

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Since both the acid and base exist in the same volume, we can skip the concentration calculations and use the number of moles in the Henderson-Hasselbalch equation to calculate the pH. The answer will be the same.

Part B

What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.

Express the pH numerically to three decimal places.

Hints

pH =

6.165

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Even though a significant amount of a strong acid was added, the pH of the buffer changed very little. If this amount of acid was added to 2.00 L of pure water, the pH would change from 7.000 to 1.125.

Part C

What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

Express the pH numerically to three decimal places.

pH =

6.408

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01:160:162 Lecture Notes - Lecture 5: Sodium Hydroxide, Equivalence Point, Titration

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