Question

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

Answer 1

Answer:

At standard room temperature, $[{\rm OH^{-}] \approx 2.5 \times 10^{-7}\; \rm M$ when $[{\rm H^{+}] = 4.0 \times 10^{-8}\; \rm M$.

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, $[{\rm H^{+}}] = 4.0 \times 10^{-8}\; \rm mol \cdot L^{-1}$. Substitute both values into the equation and solve for $[{\rm OH^{-}}]$.

$\begin{aligned} {[}{\rm OH^{-}}{]} &= \frac{K_{\rm w}}{[{\rm H^{+}}]} \\ &\approx \frac{10^{-14}}{4.0 \times 10^{-8}} = 2.5 \times 10^{-7}\end{aligned}$.

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