BUFFERS AND TITRATION OF AN AMINO ACID Essay Example

  • Category:
    Other
  • Document type:
    Assignment
  • Level:
    High School
  • Page:
    1
  • Words:
    306

AIM OF EXPERIMENT

  • Proving the buffer capacity of [HPO42-] and [H2PO4] through the employment of calibrated pH meter

  1. Using the pH values provided [a=5.9, b=6.9 and c=7.9], calculate the ratio of [H2PO4] to [HPO42-] required in producing the buffer solutions.

  2. Take 0.1 M HPO42- and 0.1M H2PO4Solutions of appropriate volumes and mix the two solutions to yield a total volume of 25ml for each buffer solutions.

  3. Use the calibrated pH meter, then measure and record the pH value of both solutions and compare with the pH value of distilled water (25ml).

  4. Then add 2.0ml of 0.1M NaOH in each buffer sample (25ml) then to the distilled water and thoroughly mix each tube.

  5. Once the alkali is added the pH of each solution is recorded.

  6. In the tabulation of the results, a column for pH change is included. Account for pH shift magnitude in reference to pH shift direction (away from or towards the pka value).

Initial pH

pH after adding 2ml of

0.1M NaOH

Change in

Moving to or away from pka

RESULTS AND DISCUSSION

The overall equation for the reaction in as follows:

HBUFFERS AND TITRATION OF AN AMINO ACID2PO4 (aq) H+ (aq) + HPO42- (aq)

  • This keeps the system at pH of around 7.4, since it lies between 7.35 and 7.45.

The dissociation of H2PO4in water gives the following equation:

HBUFFERS AND TITRATION OF AN AMINO ACID 1 2PO4¯ + H2O HPO42- + H3O+

The equilibrium for the above reaction is given by the following reaction:

[HPO42-] [H3O+]

= 2.303 C.

(Ka + [H3O+]) 2

βKa x H3O+

pKa= pH_therefore_ Ka = [H3O+]

ax. Buffer capacity= 2.303 C

([H3O+] + [H3O+]) 2

M[H3O+] x [H3O+]

Max. Buffer capacity= 2.303 C

[H3O+] 2

C= Morality = 0.1+0.1= 0.2M

Max. Buffer capacity (β) = 2.303 x 0.2 x ¼

The buffer capacity is 0.1152.

CONCLUSION

The buffer capacity obtained above (0.1152) depends on the sum of buffer concentration.