We can adjust the pH of a buffer, within limits, to bring it to some desired val

Question

We can adjust the pH of a buffer, within limits, to bring it to some desired value. For example, if Ka for the acid is 1.0 X10-5 and the amounts of acid and conjugate base are equal, the pH of the buffer solution would be 5.0. If we wish to make a buffer of pH equal to 4.5, we need to simply select volumes of the acid and conjugate base such the ratio of [acid] to [base] would make [H3O+] equal to 10-4.5 or 3.2 X 10-5 M. So to make the desired buffer we could use 320 mL of 0.10 M acid and 100 mL of 0.10 M conjugate base. Show by calculation that these volumes will give you a buffer solution with pH of 4.5

Explanation / Answer

Given that Ka =  1.0 X10-5

[conjugate base] = molarity x volume =0.1 M x 100 mL = 10 mmol

[acid ] = molarity x volume =0.1 M x 320 mL = 32 mmol

We know that

  Henderson-Hasselbalchequation is

pH = pKa + log [conjugate base]/ [acid]

   = -logKa + log [conjugate base]/ [acid]

= - log (1.0 X10-5) + log (10 mmol/32 mmol)

= 4.5

pH = 4.5

Therefore,

pH of the buffer solution = 4.5

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