JEE Advanced 2006
- Exam
- JEE Advanced
- Year
- 2006
- Subject
- Chemistry
- Chapter
- Electrochemistry
JEE Advanced 2006 · Chemistry · Electrochemistry
We have taken a saturated solution of AgBr, K_ sp is 12 × 10^ -14 ....
We have taken a saturated solution of is . If mole of are added to 1 litre of this solution, find conductivity (specific conductance) of this solution in terms of
Correct answer
55
Step-by-step solution
Suppose the solubility of in is \begin{aligned}& \underset{s \mathrm{ mol}/ \mathrm{L}}{\mathrm{AgBr}} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}\\
& \mathrm{AgNO}_{3} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{NO}}_{3}^{-} \\
& 10^{-7} \mathrm{M} \quad10^{-7} \mathrm{M} \times10^{-7} \mathrm{M}\\
& \text{ Total } \left[\mathrm{Ag}^{+}\right]=\left(s+10^{-7}\right)\mathrm{M}\\
& K_{\mathrm{sp}}\text{ of } \mathrm{AgBr}= \left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right]\\
& 12 \times10^{-14}=\left(s+10^{-7}\right)(s)=s²+10^{-7} s \\
& s²+10^{-7} s-12 \times10^{-14}=0 \\
&
\end{aligned} s=3 \frac{k}{C} \text{ or }\begin{aligned}
\therefore \quad k_{\mathrm{Br}^{-}} & =3 \times10^{-4} \times8 \times10^{-3} \mathrm{Sm}^{-1}=24 \times10^{-7} \mathrm{Sm}^{-1}\\
k_{\mathrm{Ag}^{+}} & =4 \times10^{-4} \times6 \times10^{-3} \mathrm{Sm}^{-1}=24 \times10^{-7} \mathrm{Sm}^{-1}\\
k_{\mathrm{NO}_3}^{-}} & =1 \times10^{-4} \times7 \times10^{-3}=7 \times10^{-7} \mathrm{Sm}^{-1}\\
k_{\text{Total }} & =k_{\mathrm{Br}^{-}}+k_{\mmathrm{Ag}^{+}}+k_{\mathrm{NO}_{3}}^{-}} \\
& =\left(24 \times10^{-7}+24 \times10^{-7}+7 \times10^{-7} \mathrm{Sm}^{-1}\right. \text{ ) }\\
& =55 \times10^{-7} \mathrm{Sm}^{-1}\\
& \left.=\mathbf{5 5} \mathbf{S m}^{-1} \text{ (in term of }10^{-7} \mathrm{Sm}^{-1}\right)
\end{aligned}