Question

What is the concentration of K+K+ in 0.15 MM of K2SK2S? Express your answer to one decimal place and include the appropriate units.

Answer 1

Answer : The concentration of
`K^+` ion in 0.15 M
`K_2S` is, 0.3 M

Explanation :

The given compound is,
`K_2S`

When
`K_2S` dissociates then it gives potassium ion and sulfide ion.

The balanced dissociation reaction is:

`K_2S\rightarrow 2K^++S^(2-)`

By the stoichiometry we can say that, 1 mole of
`K_2S` dissociates to give 2 moles of
`K^+` ion and 1 mole of
`S^(2-)` ion.

Or, in terms of concentration we can say that:

0.15 M of
`K_2S` dissociates to give
`2* 0.15M=0.3M` of
`K^+` ion and 0.15 M of
`S^(2-)` ion.

Thus, the concentration of
`K^+` ion in 0.15 M
`K_2S` is, 0.3 M