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

Sagot :

jacob193

Answer:

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

Explanation:

The following equilibrium goes on in water:

[tex]{\rm H_{2}O}\, (l) \rightleftharpoons {\rm H^{+}}\, (aq) + {\rm OH^{-}}\, (aq)[/tex].

The forward reaction is known as the self-ionization of water. The ionization constant of water, [tex]K_{\rm w}[/tex], gives the equilibrium position of this reaction:

[tex]K_{\rm w} = [{\rm H^{+}] \cdot [{\rm OH^{-}}][/tex].

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

[tex]\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}[/tex].

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

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