A) \[1.5\times {{10}^{-4}}\]
B) \[1.11\times {{10}^{-5}}\]
C) \[2.11\times {{10}^{-5}}\]
D) \[1.8\times {{10}^{-5}}\]
Correct Answer: B
Solution :
| [b] Idea This problem can be solved by using the concept of Kohlrausch law of independent migration by using which we can easily calculate the value of molar conductivity of crotonic acid after determining the \[{{\wedge }_{m(HC)}}\] calculate degree of dissociation \[(\alpha )\,\alpha =\frac{{{\wedge }_{m(HC)}}}{\wedge {{_{m}^{\infty }}_{(HC)}}}\] Molar conductivity of the dissociated crotonic acid. |
| \[{{\wedge }_{m(HC)}}={{\wedge }_{m(HCl)}}+{{\wedge }_{m(NaC)}}-{{\wedge }_{(m)NaCl}}\] |
| \[=(425+82-125){{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}\] |
| \[=382\,{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}\] |
| Molar conductivity of HC |
| \[{{\wedge }_{m(HC)}}=\frac{K}{C}=\frac{3.82\times {{10}^{-4}}}{0.001}\times 1000\] |
| \[=38.2{{\Omega }^{-1}}c{{m}^{2}}\,mo{{l}^{-1}}\] |
| The degree of dissociation |
| \[\alpha =\frac{{{\wedge }_{m(HC)}}}{{{\wedge }_{{{m}^{\infty }}(HC)}}}=\frac{38.2\,{{\Omega }^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}}{382\,{{\Omega }^{-1}}c{{m}^{2}}\,mo{{l}^{-1}}}=0.1\] |
| \[{{K}_{a}}=\frac{C{{\alpha }^{2}}}{1-\alpha }=\frac{{{10}^{-3}}{{(0.1)}^{2}}}{1-0.1}=1.11\times {{10}^{-5}}\] |
| TEST Edge Similar questions are asked in the question in order to determine the pH of electrolyte solution. |
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