At `25^(@)C` what is the concentration of `H_(3)O^(+)` ions in an aqueous solution in which the concentration of `OH^(-)` ions is `2xx10^(-5)(M)`?

Explanation

At `25^(@)C,K_(w)=10^(-14)`, Now, for an aqueous solution,
`[H_(3)O^(+)]xx[OH^(-)]=K_(w)`
At `25^(@)C,[H_(3)O^(+)][OH^(-)]=10^(-14)`
Given: `[OH^(-)]=2xx10^(-5)(M)`
Therefore, `[H_(3)O^(+)]=(10^(-14))/(2xx10^(-5))=5xx10^(-10)(M)`.