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8.CHEMICAL EQUILIBRIUM eLearn.Punjab

Taking the log, multiplying with negative sign and rearranging, we get

pOH = pK b + log [salt]
[base]

Using this relationship, we can prepare a basic buffer of the required pOH or pH by suitably

selecting a base and adj usting the ratio of [salt] / [base].

Example :

Calculate the pH of a buffer solution in which 0.11 molar CH3COONa and 0.09 molar acetic
acid solutions are present. Ka for CH3COOH is 1.85 x 10-5

Solution:

0.11M CH3COONa solution means that 0.11 moles are dissolved in 1 dm3 of solution.

[CH3COONa] =0.11M

[CH3COOH] =0.09M

Ka of CH3COOH =1.85x10-5

pK = -log(1.8x10-5) = 4.74

pH = pKa + log [salt]
[acid]

= pKa + log 0.11
0.09
pH
= 4.74 + 0.087 = 4.83 Answer

pH

Since, the concentration of CH3COONa is more than that of CH3COOH, so pH of buffer is
greater than 4.74. In other words, the solution has developed the properties of a base, because

CH3COONa has Na+ ion which is from a strong base.

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