Question

A solution has [H3O+] = 2.8x10-5 M. Use the ion product constant of water K. = [H30 - ) [OH-] to find the [OH-] of the solution. Express your answer to two significant figures.​

Answer 1

Approximately
`3.6 * 10^(-10)\; {\rm M}`, assuming that the solution is dilute and is at room temperature (such that
`K_\text{w} = 10^(-14)`.)

Step-by-step explanation:

Look up the ion product constant of water:

`K_{\text{w}} \approx 10^(-14)`.

In a dilute solution where water is the solvent, the product of
`[{\rm {H_(3)O}^(+)}]` and
`[{\rm {OH}^(-)}]` (concentration of
`\rm {H_(3)O}^(+)}` ions and
`{\rm {OH}^(-)}` ions) is constantly equal to
`K_{\text{w}}`. (The unit of both
`[{\rm {H_(3)O}^(+)}]\!` and
`[{\rm {OH}^(-)}]\!` need to be
`{\rm M}`.) In other words:

`[{\rm {H_(3)O}^(+)}]\, [{\rm {OH}^(-)}] = K_{\text{w}}`.

Rearrange this equation to find
`[{\rm {OH}^(-)}]` in terms of
`K_{\text{w}}` and
`[{\rm {H_(3)O}^(+)}]`:

`\begin{aligned}[] [ {\rm OH^(-)} ] &= \frac{K_{\text{w}}}{[{\rm {H_(3)O}^(+)}]} \\ &= (10^(-14))/(2.8 * 10^(-5)) \\ &\approx 3.6 * 10^(-10)\end{aligned}`.