Chemistry LabReportbuffer

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Zahoor Rahimtoola

IBD Chem HL

Chemistry Lab Report

PART 1

Aim:

  1. To make up 100 cm3 of a buffer solution of pH 5.2

To make an appropriate buffer, the ratio of the volumes of the acid and its salt needs to be calculated. To find the volumes:

The pKa value for 0.5 M ethanoic acid and 0.5 M sodium ethanoate is 4.76.

To make the buffer pH 5.2:

CH3COOH + H2O ↔ CH3COO- + H3+O

CH3COONa → CH3COO- + Na+

Using the equation previously stated:

5.2 = 4.76 – log (A-)/(HA)

10-0.44 = (A-)/(HA)

0.363 = (A-): (HA)

Therefore,

100 x 0.363/1.363 = 26.6 ± 0.1 cm3 acid (ethanoic acid) is required                        

100 – 28.46 = 73.4 cm3 ± 0.1 cm3 salt (sodium ethanoate) is required

Apparatus: 250 mL beakers, 50 cm3 burette

Method:

  1. Using a burette measure out 73.4 ± 0.1 cm3 of sodium ethanoate. (Rinse the burette with water at regular intervals)
  2. Then, using the burette measure out 26.6 ± 0.1 cm3 of ethanoic acid. Add them
  3. Then, measure the pH of the solution and compare it with the value needed.
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PART 2

Aim:

  1. How does the addition of small amounts of strong acid and strong base affect the buffer set up in part 1? And what is the effect of diluting the solution on its buffering capacity? Does it act as a better buffer when small amounts of acid are added to it or small amounts of base?

Hypothesis: The introduction of small amounts of acid or alkali to the buffer solution should have little effect on the pH of the solution due to the nature of the buffer solution. Also, the buffer should act better when ...

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