How would you write the formula of the conjugate base of the Bronstead-Lowry acids H2SO3 and HI?

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Answer 1 · 2018-07-17T02:10:06

for $H_{2} S O_{3}$ $H_2SO_3$ : the conjugate base is $H S O_{3}^{-}$ $HSO_3^-$

for $H I$ $HI$ : the conjugate base is $I^{-}$ $I^-$

Given: Bronstead-Lowry acids $H_{2} S O_{3}$ $H_2SO_3$ and $H I$ $HI$

Bronstead-Lowry acid, base equation:

$B + H A \Leftrightarrow H B^{+} + A^{-}$ $B + HA iff HB^+ + A^-$

where $B$ $B$ : base proton receiver; $H A$ $" "HA$ is the acid proton donor;

$H B^{+}$ $HB^+$ is the conjugate acid; $A^{-}$ $" "A^-$ is the conjugate base

For sulfurous acid $(H_{2} S O_{3})$ $(H_2SO_3)$ :

$H_{2} S O_{3} (a q) + H_{2} O (l) \Leftrightarrow H_{3} O^{+} (a q) + H S O_{3}^{- (a q)}$ $H_2SO_3 (aq) + H_2O (l) iff H_3O^+ (aq) + HSO_3^- (aq)$

the conjugate base is $H S O_{3}^{-}$ $HSO_3^-$

For hydroiodic acid $(H I)$ $(HI)$ :

$H I (a q) + H_{2} O (l) \Leftrightarrow H_{3} O^{+} (a q) + I^{- (a q)}$ $HI (aq) + H_2O (l) iff H_3O^+ (aq) + I^- (aq)$

the conjugate base is $I^{-}$ $I^-$