Question

Question #f1da9

Answer 1

`["OH"^- ] = 6.3xx10^-13` `M`

Explanation:

We're asked to find the hydroxide ion concentration `["OH"^(-)]`, given the hydronium ion concentration `["H"_3"O"^+]`.

At `25` `""^"o""C"`, we have the relationship

`ul(K_w = ["OH"^(-)]["H"_3"O"^+] = 1.00xx10^-14color(white)(l)M^2`

Since `["H"_3"O"^+]` is given as

`1.6xx10^-2` `M`

we plug that into the equation and solve for `["OH"^(-)]`:

`["OH"^(-)] = (1.00xx10^-14color(white)(l)M^2)/(1.6xx10^-2color(white)(l)M) = color(red)(ulbar(|stackrel(" ")(" "6.3xx10^-13color(white)(l)M" ")|)`

rounded to `2` significant figures.