# A reaction occurs when aqueous solutions of hydrosulfuric acid and potassium hydroxide are...

## Question:

A reaction occurs when aqueous solutions of hydrosulfuric acid and potassium hydroxide are combined. Write the balanced equation for this reaction.

## Neutralization Reaction:

Acid and base react to form water and salt. This is called a neutralization reaction. Hydrosulfuric acid is a weak acid, and potassium hydroxide is a strong base. Here basic salts are formed.

The reaction between aqueous solutions of hydrosulfuric acid and potassium hydroxide is a neutralization reaction in which hydrosulfuric acid is an acid and potassium hydroxide is a base.

The aqueous solution of {eq}{{\rm{H}}_{\rm{2}}}{\rm{S}} {/eq} and {eq}{\rm{KOH}} {/eq} on reaction gives potassium sulfide {eq}{{\rm{K}}_{\rm{2}}}{\rm{S}} {/eq} and water {eq}{{\rm{H}}_{\rm{2}}}{\rm{O}} {/eq} as the products. The equation for the reaction between {eq}{{\rm{H}}_{\rm{2}}}{\rm{S}} {/eq} and {eq}{\rm{KOH}} {/eq} is:

{eq}{H_2}{S_{\left( {aq} \right)}}\; + \;KO{H_{\left( {aq} \right)}}\; \to {K_2}{S_{\left( {aq} \right)}}\; + \;{H_2}{O_{\left( l \right)}} {/eq}

On balancing the above equation, we get:

The number of potassium atoms on the product side is 2, and on the reactant side, it is 1. On multiplying KOH with 2, the number of potassium atoms is balanced.

The total number of the hydrogen atoms on the reactant side is 4. Now, in order to balance the hydrogen atoms on the product side we need to multiply {eq}{{\rm{H}}_{\rm{2}}}{\rm{O}} {/eq} with 2.

Therefore, the balanced equation for the above reaction is:

{eq}{H_2}{S_{\left( {aq} \right)}}\; + \;2KO{H_{\left( {aq} \right)}}\; \to {K_2}{S_{\left( {aq} \right)}}\; + \;2{H_2}{O_{\left( l \right)}} {/eq}