BCH 2333 Lecture Notes - Lecture 3: Cystine, Glutamine, Absolute Configuration

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17 Jan 2017

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Can be calculated by the henderson=hasselbach equation. As oh- is added to the reaction, it reacts completely with ha to form a- X = the equivalents of oh- added and v represesnts the volume of the solution co represents. ** we can get pkas from titration curves. We can incorporate this into the henderson-hasselbalch equation. Resist change in ph even if you are adding different acids/bases are added to the solution. Very useful in maintaining ph essential in biochemistry. Typically they are composed of a weak acid and its conjugate base. Acids and their conjugate bases are in equilibrium equilibria are related to the properties of reactants and products, so for weak acids, the tendency to give up its proton determines its buffering property. The tendency to ionize can be put in an equilibrium equation: A solution of weak acid that has a ph near its pka has equivalent amounts of conjugate base and weak acid.

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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

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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

BCH 2333 Lecture Notes - Lecture 3: Cystine, Glutamine, Absolute Configuration

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