CHM 145 Lecture 15: CHM 145 2/28/17

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8 Mar 2017

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Common-ion effect: the common-ion effect is the shift in the position of an equilibrium caused by the addition of an ion taking part in the reaction. Example: combining solutions of naf and hf hf (aq) + H2o(l) <==> h3o+ (aq) + f (aq) the presence of the additional ion reduces the degree of ionization (i. e. , shifts left): addition of strong acid (h3o+ ) or base (oh ) can also act as a. Common ion for a weak acid or weak base. Why 2nd step of diprotic acid ionization is small! The henderson-hasselbalch equation: a variation of the ionization constant expression, consider a hypothetical weak acid, ha, and its salt, only useful when you can use initial concentrations of acid and salt. This limits the validity of the equation (i. e. , change due to x must be small). Calculate the ph of a solution containing 0. 22 m acetic acid (hc2h3o2 or ch3cooh) and 0. 13. 0. 22 m acetic acid (hc2h3o2) = weak acid.

Related Questions

Need help with part B and C

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

SubmitMy AnswersGive Up

Correct

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

All attempts used; correct answer displayed

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

Incorrect; Try Again; 4 attempts remaining

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

CHM 145 Lecture 15: CHM 145 2/28/17

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