[Blueberries116]

The problem is as follows:

In a galvanic cell the cathode is an [itex]Ag^{+}(1.00\,M)/Ag_{(s)}[/itex] half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing [itex]0.10\,M[/itex] benzoic acid [itex](C_6H_5COOH)[/itex] and [itex]0.050\,M[/itex] of sodium benzoate [itex](C_6H_5COO^{-}Na^{+})[/itex]. The measured cell voltage is [itex]1.030\,V[/itex]. What is the [itex]pK_a[/itex] of benzoic acid?.

What I did to solve this problem was to find the potential for the cell involving the standard hydrogen electrode. At first I was confused because there are three "elements" featured in the problem. One being the silver electrode, the other the standard hydrogen electrode and the other a buffer solution, so I didn't know how to proceed from there. Then I noticed that to get the constant of equilibrium I only require the concentration of [itex][H^{+}][/itex] ions as,

[itex]K_a=\frac{[C_6H_5COO^{-}][H^{+}]}{[C_6H_5COOH]}[/itex]

Therefore to obtain those protons I did this:

The half equations in the cell are:

[itex]\begin{array}{cc}
Ag^{+}+1e^{-}\rightarrow Ag_{(s)}&E^{0}=0.7999\,V\\
H^{+}+1e^{-}\rightarrow \frac{1}{2}H_{2(g)}&E^{0}=0.0000\,V\\
\end{array}[/itex]

Hence the overall reaction for this process would be:

[itex]E^{0}_{cell}=E_{cathode}-E_{anode}=0.7999-0.0000=0.7999\,V[/itex]

Which is for:

[itex]Ag^{+}+\frac{1}{2}H_{2}+\rightarrow Ag_{(s)} + H^{+}[/itex]

Hence:

[itex]E_{cell}=E^{0}-\frac{0.0592}{n}\log\frac{[H^{+}]}{[Ag^{+}]p^{\frac{1}{2}}_{H_{2(g)}}}[/itex]

Since it indicates that the cell potential is [itex]1.030\,V[/itex] then:

[itex]1.030=0.7999-\frac{0.0592}{1}\log\frac{[H^{+}]}{[1](1)^{\frac{1}{2}}}[/itex]

Solving this I'm getting:

[itex][H^{+}]=0.000125306\,M[/itex]

Now all that's left is to plug in this value in the equation to get the equilibrium constant:

[itex]K_a=\frac{[C_6H_5COO^{-}][H^{+}]}{[C_6H_5COOH]}[/itex]

[itex][C_6H_5COO^{-}]=[H^{+}]=0.05\,M and [C_6H_5COOH]=0.1\,M[/itex]

Hence:

[itex]K_a=\frac{(0.05)(1.25306\times 10^{-4})}{(0.1)}=6.2653\times 10^{-5}[/itex]

Therefore the $pKa$ of benzoic acid would be:

[itex]pKa=-\log Ka=-\log\left(6.2653\times 10^{-5}\right)=4.20306[/itex]

Which does seem to be within the value of benzoic acid which I have on different references. But the problem with this method it is that it required the use of logarithm.

Given this situation, does it exist an approximation or anything that can be done right of the bat to get an idea where that value would be?. Does it exist another method which I could use?.

[Borek]

No way to avoid use of the logarithm here and your approach looks OK to me.

Broadly speaking you can express pKa using pH and log of ratio of concentrations of the acid and conjugate base. While pH can be directly calculated from the Nernst equation (it contains the term log(H⁺), there is no way to avoid finding log of the acid/base part.