For a tribasic acid, H3A, Ka1 = 2 × 10−5, Ka2 = 5 × 10−9 and Ka3 = 4 × 10−12. The value of $\frac{\l

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For a tribasic acid, H3A, Ka1 = 2 × 10−5, Ka2 = 5 × 10−9 and Ka3 = 4 × 10−12. The value of $\frac{\left\lbrack A^{3 -} \right\rbrack}{\left\lbrack H_{3}A \right\rbrack}$at equilibrium in an aqueous solution originally having 0.2 M – H3A is

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$\left\lbrack H^{+} ightbrack = \sqrt{0.2 	imes 2 	imes 10^{- 5}} = 2 	imes 10^{- 3}	ext{ M}$Now, $\left( 2 	imes 10^{- 5} ight) 	imes \left( 5 	imes 10^{- 9} ight) 	imes \left( 4 	imes 10^{- 12} ight) = \frac{\left( 2 	imes 10^{- 3} ight)^{3} 	imes \left\lbrack A^{3 +} ightbrack}{\left\lbrack H_{3}A ightbrack}$$	herefore\frac{\left\lbrack A^{3 -} ightbrack}{\left\lbrack H_{3}A ightbrack} = 5 	imes 10^{- 17}$

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