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# Write the conjugate acid of $HS$.

Last updated date: 17th Jun 2024
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Answer
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Hint: Conjugate acid is an acid which is formed by a base by accepting a proton from an acid. Both of them form a conjugate acid-base pair.

Complete answer:
According to the Bronstead-Lowry concept an acid is a substance which donates protons and a base is a substance which accepts protons. The donation of proton by the acid generates a conjugate base and the acceptance of proton by the base generates a conjugate acid.
Let us consider the dissolution of hydrochloric acid in water. The reaction for the dissociation of $HCl$ in water is:
$HCl + {H_2}O \to {H_3}{O^ + } + C{l^ - }$.
The reaction indicates that the water accepts protons from the acid to generate hydronium ion (${H_3}{O^ + }$). Thus water acts as a base in this reaction and the product formed by addition of proton is the corresponding conjugate acid.
The question asks to write the conjugate acid of $HS$. $HS$ exists as hydrosulphite ion $H{S^ - }$. As seen earlier conjugate acid is the acid obtained after addition of a proton to a base. So the conjugate acid of $H{S^ - }$ can be written as:
$H{S^ - } + {H_2}O \to {H_2}S + O{H^ - }$
From the above equation it is seen that the hydrosulfite ion accepts a proton from water. Thus it acts as a base and water as an acid. The corresponding conjugate acid of $H{S^ - }$ formed by addition of proton is hydrosulfuric acid i.e.${H_2}S$.

Note:
The count of the number of hydrogens on each substance before and after the reaction indicates whether a substance is an acid or a base, simply. In case the number of hydrogens has decreased then the substance is the acid. In case if the number of hydrogens has increased then the substance is the base. The substances on the left side of the equation are called acid and base while on the right side are called conjugate acid and conjugate base. The original acid gets turned into the conjugate base after the reaction.