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

Question

8) 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 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 that the ratio of [acid] to [base] would make H,O] equal to 10 or 3.2 X 10 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 4.5 9) For the same acid and base in (8), what ratio of volumes of acid and base would you need to get a pH of 5.5? Show your calculation. Give a set of volumes that satisfies the calculated ratio

Explanation / Answer

Not to worry! It's quite simple if you follow the steps shown below.

First of all, you have to understand how the calculations were done for getting pH = 4.5, then you can easily perform the calculations for getting pH = 5.5.

Calculations for getting the pH = 4.5

According to Henderson-Hasselbulch equation:

pH = pKa + Log{(M*V)conjugate base/(M*V)acid}

Where M is the corresponding molar concentration and V is the corresponding volume in mL

Here, Mconjugate base = Macid

And pKa = -Log(Ka) = -Log(10-5) = 5

Therefore, pH = 5 + Log(Vconjugate base/Vacid)

i.e. pH = 5 + Log(100 mL/320 mL)

Here, Log(100/320) = -0.5

i.e. pH = 5 - 0.5 = 4.5

Calculations for getting the pH = 5.5

pH = 5 + Log(Vconjugate base/Vacid)

In the previous case for getting pH = 4.5, Log(Vconjugate base/Vacid) = -0.5

Now, if Log(Vconjugate base/Vacid) = +0.5, then you will get pH = 5.5

So, to make the logarithm value = +0.5, you need to interchange the volumes of acid and conjugate base. The logic is that Log(1/x) = Log(x-1) = -Logx

i.e. The volume of acid = 100 mL and the volume of conugate base = 320 mL

Therefore, the volumes of acid and base needed to get pH = 5.5 is 100 : 320 = 1 : 3.2

Is it clear now?

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