Question

AA solution contains 0.09 M HCl , 0.09 M Cl2CHCOOH and 0.1 M CH3COOH . If the pH of above solution is 1 . Ka for acetic acid is 10-5 . Determine Ka for Cl2CHCOOH.

Answer 1

$For\quad a\quad weak\quad acid,\quad the\quad experssion\quad for\quad the\quad hydrogen\quad ion\quad concentration\quad is$

$[{ H }^{ + }]\quad =\quad \sqrt { c\quad \times \quad { K }_{ a } }$

$The\quad total\quad hydrogen\quad ion\quad concentration\quad of\quad the\quad solution\quad will\quad be,$

$[{ H }^{ + }{ ] }_{ total }\quad =\quad [{ H }^{ + }{ ] }_{ HCl }\quad +\quad [{ H }^{ + }{ ] }_{ CH{ Cl }_{ 2 }COOH }\quad [{ H }^{ + }{ ] }_{ C{ H }_{ 3 }COOH }$

$From\quad the\quad given,\\ molarity\quad of\quad HCl\quad =\quad 0.09\quad M$

$molarity\quad of\quad C{ H }_{ 3 }COOH\quad =\quad 0.1\quad M\\ molarity\quad of\quad { Cl }_{ 2 }CHCOOH\quad =\quad 0.09\quad M$

$[{ H }^{ + }{ ] }_{ total }\quad =\quad 0.1\quad =\quad 0.09\quad +\quad \sqrt { { c }_{ CHCl_{ 2 }COOH }\quad \times \quad { K }_{ aCHCl_{ 2 }COOH } } \quad +\quad \sqrt { c_{ C{ H }_{ 3 }COOH }\quad \times \quad { K }_{ aC{ H }_{ 3 }COOH } }$

$0.01\quad =\quad \sqrt { \quad 0.09\quad \times \quad { K }_{ aCHCl_{ 2 }COOH } } \quad +\quad \sqrt { 0.1\quad \times \quad { 10 }^{ -5 } }$

$\quad \quad { K }_{ aCHCl_{ 2 }COOH }\quad =\quad 0.9\quad \times \quad { 10 }^{ -4 }$