CHEM 1062 Lecture Notes - Lecture 19: Buffer Solution, Ammonium Chloride, Conjugate Acid

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13 Nov 2016

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Buffer - if you do something to the system, the system will counter to balance out the change ph buffer - resistant to the change to the system, will maintain the ph of the solution. If you add h3o+ then the reaction will move to the left (eq shifts left) to use up added. H3o+ to make ha & h2o, a- is also used up. If you add oh-, it will react with h3o+ so h3o+ will decrease, system will move to the right to make more; ph stays constant; a- also produced (ha &h2o used up) Adding water (diluting the solution), concentration of h3o+ decreases; system will move to the right to increase h3o+ concentration (producing a- and using up ha & h2o) In order to have a buffer, need to have a weak acid and its weak conjugate base in solution at similar concentrations.

Related Questions

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

Need help with part B and C

1.) Include a brief (fewer than 4sentence) explanation for each of the following that cites raw data or information gathered from plots:

a. In both words and in terms of a mathematical equation, describe the relationship between pH and the ratio of conjugate base to acid concentration that you observed in this experiment. Compare the plots of [conj. base]/[conj. acid] vs. pH and log([conj. base]/[conj. acid]) vs. pH

b. Compare the addition of a strong acid to water and the addition of a strong acid to your buffer. Suggest a reason for any difference you observe.

c. Describe the effect of diluting the acid. Upon sufficient dilution, what would you predict the final pH to be?

d. Compare the pH values for your most diluted buffer sample to the undiluted sample. Do the same for your acid and base samples. Do you notice any difference between the buffer and the acid or base? Suggest a reason for any difference you see.

2.) A common biological buffer is a âphosphate buffer,â which involves the equilibrium between H2PO4^- and HPO4^(2)-, with a pKa at 25C of 6.86. Describe in words how you would prepare 500.mL of 0.250 M phosphate buffer at pH = 7.2, given appropriate glassware and 500. mL each of 0.250 M NaH2PO4 and 0.250 M Na2HPO4 at 25C. (Note: your explanation needs to include actual volumes, so this question requires calculations! It will be helpful to know that the concentrations of each component must add to the buffer concentration. i.e. [A-] + [HA] = 0.250 M)

CHEM 1062 Lecture Notes - Lecture 19: Buffer Solution, Ammonium Chloride, Conjugate Acid

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