Question

What is the OH- of {H+} = 4.0 x 10 to the power of -8

Answer 1

At standard room temperature,
`[{\rm OH^(-)] \approx 2.5 * 10^(-7)\; \rm M` when
`[{\rm H^(+)] = 4.0 * 10^(-8)\; \rm M`.

Step-by-step explanation:

The following equilibrium goes on in water:

`{\rm H_(2)O}\, (l) \rightleftharpoons {\rm H^(+)}\, (aq) + {\rm OH^(-)}\, (aq)`.

The forward reaction is known as the self-ionization of water. The ionization constant of water,
`K_(\rm w)`, gives the equilibrium position of this reaction:

`K_(\rm w) = [{\rm H^(+)] \cdot [{\rm OH^(-)}]`.

At standard room temperature (
`25\; {\rm ^(\circ)C}`),
`K_(\rm w) \approx 10^(-14)`. Also,
`[{\rm H^(+)}] = 4.0 * 10^(-8)\; \rm mol \cdot L^(-1)`. Substitute both values into the equation and solve for
`[{\rm OH^(-)}]`.

`\begin{aligned} {[}{\rm OH^(-)}{]} &= \frac{K_(\rm w)}{[{\rm H^(+)}]} \\ &\approx (10^(-14))/(4.0 * 10^(-8)) = 2.5 * 10^(-7)\end{aligned}`.

In other words, in an aqueous solution at standard room temperature,
`[{\rm OH^(-)] \approx 2.5 * 10^(-7)\; \rm M` when
`[{\rm H^(+)] = 4.0 * 10^(-8)\; \rm M`.