The molar conductivity of a solution of a wea | vedclass

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Question

$HX \rightleftharpoons H ^{+}+ X ^{-}$

$Ka =\frac{\left[ H ^{+}\right]\left[ X ^{-}\right]}{[ HX ]}$

$HY \rightleftharpoons H ^{+}+ Y ^{-}$

$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$HX ightleftharpoons H ^{+}+ X ^{-}$

$Ka =\frac{\left[ H ^{+}ight]\left[ X ^{-}ight]}{[ HX ]}$

$HY ightleftharpoons H ^{+}+ Y ^{-}$

$Ka =\frac{\left[ H ^{+}ight]\left[ Y ^{-}ight]}{[ HY ]}$

$\Lambda_{ m } 	ext { for } HX =\Lambda_{ m _2}$

$\Lambda_{ m } 	ext { for } HY =\Lambda_{ m _2}$

$\Lambda_{ m _1}=\frac{1}{10} \Lambda_{ m _2}$

$Ka _2= C ^2$

$Ka _1= C _1 	imes\left(\frac{\Lambda_{ m _1}}{\Lambda_{ m _1}^0}ight)^2$

$Ka _2= C _2 	imes\left(\frac{\Lambda_{ m _2}}{\Lambda_{ m _2}^0}ight)^2$

$\frac{ Ka _1}{ Ka _2}=\frac{ C _1}{ C _2} 	imes\left(\frac{\Lambda_{ m _1}}{\Lambda_{ m _2}}ight)^2=\frac{0.01}{0.1} 	imes\left(\frac{1}{10}ight)^2=0.001$

$pKa _1- pKa _2=3$

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