Question

\(1 \times 10^{-5} M \,AgNO _3\)is added to \(1\, L\) of saturated solution of\(AgBr\) The conductivity of this solution at \(298 \,K\) is ___ [Given : \(K _{\text {SP }}( AgBr )=49 \times 10^{-13} at 298\, K\)\(\lambda_{ Ag ^{+}}^0=6 \times 10^{-3} S\, m ^2\, mol ^{-1}\) \(\lambda_{ Br ^{-}}^0=8 \times 10^{-3} S\, m ^2\, mol ^{-1}\) \(\lambda_{ NO _3^{-}}^0=7 \times 10^{-3} \,S\, m ^2 \,mol ^{-1}\) ]

Hint:

Remember the relationship between conductivity, molar conductivity, and concentration: \(κ =\sumλ_ic_i\). Also, be mindful of the common ion effect when dealing with solubility equilibria.

Answer 1

Correct Answer: 14

The dissociation of AgBr is:
\[\text{AgBr}(s) \leftrightharpoons \text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}),\]
\[K_{\text{sp}} = [\text{Ag}^+][\text{Br}^-] = 4.9 \times 10^{-13}.\]
Let the solubility of AgBr be \(s\) mol/L. Then, \([\text{Ag}^+] = s\) and \([\text{Br}^-] = s\).
\[s^2 = 4.9 \times 10^{-13}, \quad s = \sqrt{4.9 \times 10^{-13}} = 7 \times 10^{-7} \, \text{M}.\]
Since \(1 \, \text{L}\) of saturated AgBr solution is taken, the concentration of \(\text{Ag}^+\) and \(\text{Br}^-\) from AgBr are both \(7 \times 10^{-7} \, \text{M}\). We are adding \(1 \times 10^{-5} \, \text{M}\) AgNO\(_3\).
The \(\text{Ag}^+\) from AgNO\(_3\) will be significantly greater than the \(\text{Ag}^+\) from AgBr, so we can approximate the total \([\text{Ag}^+]\) as \(1 \times 10^{-5} \, \text{M}\).
The common ion effect will suppress the solubility of AgBr, so the \([\text{Br}^-]\) remains approximately \(7 \times 10^{-7} \, \text{M}\). The \([\text{NO}_3^-]\) will be \(1 \times 10^{-5} \, \text{M}\).
Conductivity (\(\kappa\))
The formula for conductivity is:
\[\kappa = \sum \lambda_i c_i.\]
Substitute the values:
\[\kappa = \lambda_{\text{Ag}^+}[\text{Ag}^+] + \lambda_{\text{Br}^-}[\text{Br}^-] + \lambda_{\text{NO}_3^-}[\text{NO}_3^-].\]
\[\kappa = (6 \times 10^{-3})(1 \times 10^{-5}) + (8 \times 10^{-3})(7 \times 10^{-7}) + (7 \times 10^{-3})(1 \times 10^{-5}).\]
\[\kappa = 6 \times 10^{-8} + 5.6 \times 10^{-9} + 7 \times 10^{-8}.\]
\[\kappa \approx 13.56 \times 10^{-8} = 14 \times 10^{-8} \, \text{S m}^{-1}.\]
Final Answer:
The conductivity is \(14 \times 10^{-8} \, \text{S m}^{-1}\).