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a buffer solution contain 0.1 mole per litres of acetic acid and 0.001 moles perlitre of sodium acetate.what will be its pH?(k=1.8×10-5)

Sagot :

Answer: The pH of the solution is 2.74.

Explanation:

To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:

[tex]pH=pK_a+\log (\frac{\text{[conjugate base]}}{\text{[acid]}})[/tex]         .......(1)

The power of the acid dissociation constant is the negative logarithm of the acid dissociation constant. The equation for it is:

[tex]pK_a=-log K_a[/tex]        ......(2)

The chemical equation for the reaction of acetic acid and NaOH follows:

[tex]CH_3COOH+NaOH\rightleftharpoons CH_3COONa+H-2O[/tex]

Given values:

[tex]K_a=1.8\times 10^{-5}[/tex]

Putting values in equation 2:

[tex]pK_a=-log (1.8\times 10^{-5})\\\\pK_a=4.74[/tex]

We are given:

[tex]CH_3COOH[/tex] = 0.1 M

[tex]CH_3COONa[/tex] = 0.001 M

Plugging values in equation 1:

[tex]pH=4.74+\log (\frac{0.001}{0.1})\\\\pH=4.74-2\\\\pH=2.74[/tex]

Hence, the pH of the solution is 2.74.

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