Question

A sample of lead weighing $1.05g$ was dissolved in a small quantity of nitric acid to produce an aqueous solution of ${Pb}^{2+}$ and ${Ag}^{+}$ (which is present as impurity). The volume of the solution was increased to $300ml$ by adding water, a pure silver electrode was immersed in the solution and the potential difference between this electrode and a standard hydrogen electrode was found to be $0.503V$ at ${25}^{o}C$. What was the % of $Ag$ in the lead metal? (Given : $E_{\left({Ag}^{+}/Ag\right)}^o=0.799V$. Neglect amount of ${Ag}^{+}$ converted to $Ag$)

Answer 1

Solution:

The mass fraction of silver is :

$\omega =\frac{m\left(Ag\right)}{m\left(\text{sample}\right)}$

where m(sample) is 1.05g . Lets find the concentration of silver in solution. For this, we can use Nernst equation:

$E=E^0+0.0591\times ln\left[A{g}^{+}\right]$

$0.503=0.799+0.0591\times ln\left[A{g}^{+}\right]$

$ln\left[A{g}^{+}\right]=-5.01$

$\left[A{g}^{+}\right]=6.68\times {10}^{-3}mol{L}^{-1}$

The number of the mioles of $A{g}^{+}$ in 300ml solution is :

$n\left(Ag\right)=\left[{Ag}^{+}\right]\times V=6.68\times {10}^{-3}\times 300\times {10}^{-3}=2.00\times {10}^{-3}$mol

The mass of the silver is :

$m\left(Ag\right)=n\left(Ag\right)\times M\left(Ag\right)=2.00\times {10}^{-3}\times 107.8682=0.216g$

Mass fraction of the silver is :

$\omega =\frac{0.216}{1.05}=0.21or21$%

Hence answer is 21% of Ag