Determination of pKa value by Half Neutralization/ Henderson Hasselbalch equation

BACKGROUND

Aim: 

To determine the pKa value of a weak acid by titrating it against strong base potentiometrically (using a pH meter).

When acetic acid is titrated against strong base it forms a mixture of acid and its salt. The mixture acts as a buffer solution. During the course of titration acid concentration decreases and salt concentration goes on increasing. During any stage of titration the pH of buffer solution is explained by the Henderson –Hasselbalch equation.¹

pH = pK_a + log [Salt]/ [Acid] …… (1)

Where, pKa = -logKa and Ka = dissociation constant of the acid

[Salt] = concentration of salt CH₃COONa

[Acid] = concentration of acid CH₃COOH

At the half neutralization point [Salt] = [Acid]

And equation (1) above gives pH = pKa

REQUIREMENTS

Chemicals:    

0.04N acetic acid

0.04N NaOH (Standard)

Buffer solution of known pH

Apparatus:     

Burette

Beakers

Stirrer

pH meter

PROCEDURE

Standardise the pH meter by the use of a buffer solution of known pH. Pipette out 25ml of 0.04N acetic acid solution in a dry beaker. Dip the electrode into the acid. Measure the pH. Add from the burette 1ml of 0.04N NaOH and stir. Measure the pH (do not remove the glass rod used for stirring). Continue addition of 1ml of NaOH till there is sudden change in pH and thereafter take at least three more readings. After each addition measure the pH.

Observations

Room temperature = ^◦C

V ml of 0.04N NaOH	pH	dpH	dV	dpH/dV	Total volume ml	dB	β= dB/dpH
1
2
3
4

Graph

• Plot the graph of pH vs volume of 0.04N NaOH ‘V’
  
• To determine exact end point plot dpH/dV vs V
  
• Plot a graph of β vs pH. Find the pH at which β is maximum.

Calculation

Read the end point of the titration from graph (2). Let it be xml of 0.04N NaOH: Now from graph (1) read the pH for V = x/2, i.e. read the pH at half neutralization stage at which [Salt] = [Acid] and from equation (1) pH = pK_a value,

Therefore K_a =

Calculation of dB:

Since, 1000 ml of 1N NaOH = 1 equivalent of NaOH

Therefore 2 ml of 0.04 N NaOH = (2 ×1×0.04)/(25 x 1000) equivalent of NaOH

If ‘x’ ml of 0.04N NaOH is added to 25ml of 0.04N acetic acid solution, the total volume of the solution is (25+x)ml and gram equivalent of NaOH per litre of solution added per 2 ml increment in NaOH is,

dB = (2×1000) / [(25×1000) (25+x)] = 2 / 25 × total volume

CONCLUSION

The buffer capacity (β_max) is ______

pK_a value of acetic acid =______ at room temperature

K_a for acetic acid = ______ at room temperature.

REFERENCES

1. More HN, Hajare AA. Practical Physical Pharmacy, Career publications, 2010: 224-226.

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