Question

The following cell was found to have a potential of -0.492 V:

$\mathrm{Ag}\overline{)\mathrm{AgC}}\mathrm{l}\left(\mathrm{sat}’\mathrm{d}\right)\mathrm{IIHA}\left(0.200\mathrm{M}\right),\mathrm{NaA}\left(0.300\mathrm{M}\right)\overline{){\mathrm{H}}_2}\left(1.00\mathrm{atm}\right),\mathrm{Pt}$

Calculate the dissociation constant of HA, neglecting the junction potential.

Short answer

The dissociation constant for HA is $1.69\times {10}^{-5}$.

Step-by-step solution

Given information

The given cell is as follows:

$\mathrm{Ag}\overline{)\mathrm{AgC}}\mathrm{l}\left(\mathrm{sat}’\mathrm{d}\right)\mathrm{IIHA}\left(0.200\mathrm{M}\right),\mathrm{NaA}\left(0.300\mathrm{M}\right)\overline{){\mathrm{H}}_2}\left(1.00\mathrm{atm}\right),\mathrm{Pt}$

The cell potential is $-0.492\mathrm{V}$.

Half-cell reactions

In the given cell, standard hydrogen electrode is used as reference electrode.

The half-cell reaction involved in the given cell are shown below:

$$\begin{gathered} \mathrm{Cathode}:{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\left(\mathrm{g}\right) \\ \mathrm{Anode}:2\mathrm{Ag}\left(\mathrm{s}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to 2\mathrm{AgCl}\left(\mathrm{s}\right)+2{\mathrm{e}}^{-} \end{gathered}$$

Therefore, the reduction potential of anodize and cathodic half-cell reaction are 0.199 V and 0.00 V respectively.

Step 3: Ecello cell expression

The $E_{cell}^o$for the reaction is as follows:

role="math" localid="1649430583593" $E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Substitute reduction potential for half-cell reaction in above equation as follows:

$$\begin{gathered} E_{cell}^o=E_{cathode}^o-E_{anode}^o \\ =0.00\mathrm{V}-0.199\mathrm{V} \\ =-0.199\mathrm{V} \end{gathered}$$

Nernst equation 

The overall cell reaction is as follows:

$2\mathrm{Ag}\left(\mathrm{s}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to 2\mathrm{AgCl}\left(\mathrm{s}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$

The Nernst equation for the above reaction is represented as shown below:

$E_{cell}=E_{cell}^o-\frac{0.0592}{2}\log \frac{{\left[\mathrm{AgCl}\right]}^2\left[{\mathrm{H}}_2\right]}{\left[{\mathrm{Cl}}^{-}\right]\left[{\mathrm{H}}^{+}\right]{\left[\mathrm{Ag}\right]}^2}.......\left(1\right)$

Expression concentration of hydrogen ions

It is assumed that the activity of the solid and liquid substance are con taken as 1.

Substitute the concentration of each substance in above equation (1) as follows:

$$\begin{gathered} E_{cell}=E_{cell}^o-\frac{0.0592}{2}\log \frac{{\left[1\right]}^2\left[1\right]}{\left[1\right]\left[{\mathrm{H}}^{+}\right]{\left[1\right]}^2} \\ =E_{cell}^o-\frac{0.0592}{2}\log \frac{1}{\left[{\mathrm{H}}^{+}\right]} \\ =E_{cell}^o-\frac{0.0592}{2}\left(-2\times \log \left[{\mathrm{H}}^{+}\right]\right) \\ =E_{cell}^o-0.0592\times \log \left[{\mathrm{H}}^{+}\right] \\ \log \left[{\mathrm{H}}^{+}\right]=\frac{E_{cell}-E_{cell}^o}{0.0592}..........\left(2\right) \end{gathered}$$

Calculate concentration of hydrogen ion 

Substitute values in the above equation (2) as follows:

$$\begin{gathered} \log \left[{\mathrm{H}}^{+}\right]=\frac{-0.492-\left(-0.199\right)}{0.0592} \\ \log \left[{\mathrm{H}}^{+}\right]=\frac{-0.492+0.199}{0.0592} \\ \log \left[{\mathrm{H}}^{+}\right]=-4.95 \\ \left[{\mathrm{H}}^{+}\right]=1.124\times {10}^{-5}\mathrm{M} \end{gathered}$$

Dissociation of HA 

The dissociation of HA is expressed as follows:

$\mathrm{HA}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{-}$

The dissociation constant for above reaction is given as:

role="math" localid="1649661119379" $K_a=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\mathrm{HA}}........\left(3\right)$

The concentration of $\mathrm{HA}$and data-custom-editor="chemistry" ${\mathrm{A}}^{-}$are given as 0.200 M and 0.300 M respectively.

Substitute the concentrations in above equation as follows:

data-custom-editor="chemistry" $$\begin{gathered} K_a=\frac{1.122\times {10}^{-5}\times 0.300}{0.200} \\ =1.69\times {10}^{-5} \end{gathered}$$