Question

A 1.41 L buffer solution consists of 0.149 M butanoic acid and 0.288 M sodium butanoate. Calculate the pH of the solution following the addition of 0.061 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52×10−5 .

Answer 1

`pH=5.31`

Step-by-step explanation:

Hello,

In this case, by considering the Henderson-Hasselbach equation:

`pH=pKa+log(([base])/([acid]) )`

We can compute the pH before the addition of the NaOH:

`pH=-log(1.52x10^(-5))+log(([0.288M)/(0.149M) )\\\\pH=5.10`

Nevertheless, if 0.061 moles of NaOH are added, we first need to compute the present moles of butanoic acid and sodium butanoate:

`n_(acid)=0.149mol/L*1.41L=0.210mol\\\\n_(base)=0.288mol/L*1.41L=0.406mol`

So the moles of acid and base after the addition are:

`n_(acid)=0.210mol-0.061mol=0.149mol\\\\n_(base)=0.406mol+0.061mol=0.467mol`

And the concentrations in the same volume:

`[acid]=(0.149mol)/(1.41L) =0.106M`

`[base]=(0.467mol)/(1.41L) =0.331M`

Thus, the new pH is:

`pH=-log(1.52x10^(-5))+log(([0.331M)/(0.106M) )\\\\pH=5.31`

Which is a difference of pH of 0.21.

Best regards.