Question

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

$Ag\left|AgCl\right(Sat\text{'}d\left)\right|\left|HA\right(0.200M),NaA(0.300M\left)\right|H_2(1.00atm),Pt$

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

Short answer

Dissociation constant of $HA$neglecting the junction potential is $1.69\times {10}^{-5}$

Step-by-step solution

The potential of the cell.

The following equation gives the electrode potential:

$E=E^0-\frac{0.0592}{n}\log \frac{{\left(a_R\right)}^r{\left(a_S\right)}^s\dots }{{\left(a_P\right)}^p{\left(a_Q\right)}^q\dots }$ $\left(1\right)$

where ${\left(a_R\right)}^r$, ${\left(a_S\right)}^s$are activities of reactants in chemical reactions, ${\left(a_P\right)}^p$, ${\left(a_Q\right)}^q$are products in chemical reactions, exponentiated with corresponding stechiometric coefficients, and $E$is electrode potential and $E^0$is standard electrode potential.

Molar concentration is used to approximate the activity of reactants and products in cell reactions.

The equation determines the cell's potential.

$E_{cell}=E_{right}-E_{left}$ $\left(2\right)$

where $E_{Right}$denotes the potential of the hall-cell on the right in the cell diagram, and$E_{left}$ denotes the potential of the hall-cell on the left in the cell diagram.

Find the left half-cell.

Given:

$Ag\left|AgCl\right(Sat\text{'}d\left)\right|\left|HA\right(0.200M),NaA(0.300M\left)\right|H_2(1.00atm),Pt$

$E_{cell}=-0.492V$

Left half-cell,

We know from appendix three that,

$AgC{l}_{\left(s\right)}+e^{-}=A{g}_{\left(s\right)}+C{l}^{-}$

$E^0==0.199V,t={25}^{\circ }C$

$E_{left}={E^0}_{AgCl}=0.199V$

Find the Right half-cell.

$2H^{+}+2e^{-}=H_{\left(2g\right)}$

$E^0=0.000V,t={25}^{\circ }C$

Substituting the known values in $\left(1\right)$, we get

$E=E^0-\frac{0.0592}{2}\log \frac{p_{H_2}}{{\left[{\mathrm{H}}^{+}\right]}^2}$

$=0.000V-\frac{0.0592}{2}\log \frac{(1.00atm)}{{\left[H^{+}\right]}^2}$ $\left(3\right)$

Find the dissociation constant for HA.

$E_{cell}=E_{Right}-E_{left}$

Substitute the values,

$E_{cell}=E_{H^{+}/H_2}-E_{\mathrm{AgCl}}$

$-0.492\mathrm{V}=0.000\mathrm{V}-\frac{0.0592}{2}\log \frac{1.00\mathrm{atm}}{{\left[{\mathrm{H}}^{+}\right]}^2}-0.199\mathrm{V}$

To find $\left[H^{+}\right]$:

$-0.492\mathrm{V}+0.199\mathrm{V}=0.000\mathrm{V}-\frac{0.0592}{2}\log \frac{1.00\mathrm{atm}}{{\left[{\mathrm{H}}^{+}\right]}^2}$

$-0.293\mathrm{V}=0.000\mathrm{V}-\frac{0.0592}{2}\log \frac{1.00\mathrm{atm}}{{\left[{\mathrm{H}}^{+}\right]}^2}$

$-\log \left[H^{+}\right]=4.95$

$\left[H^{+}\right]=1.1238\times {10}^{-5}M$

For the species $HA$and the related reaction, the association constant is:

$HA=H^{+}+A^{-}$,

$K_a=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$

To find $K_a$:

$$\begin{gathered} K_a=\frac{1.1238\times {10}^{-5}\mathrm{M}\times 0.300\mathrm{M}}{0.200\mathrm{M}} \\ =1.69\times {10}^{-5} \end{gathered}$$