Answers: 3
![[{\rm OH^{-}] \approx 2.5 \times 10^{-7}\; \rm M](/tpl/images/2451/4808/cf07f.png)
when ![[{\rm H^{+}] = 4.0 \times 10^{-8}\; \rm M](/tpl/images/2451/4808/610b0.png)
. 
.
, gives the equilibrium position of this reaction:![K_{\rm w} = [{\rm H^{+}] \cdot [{\rm OH^{-}}]](/tpl/images/2451/4808/c45ee.png)
.
), 
. Also, ![[{\rm H^{+}}] = 4.0 \times 10^{-8}\; \rm mol \cdot L^{-1}](/tpl/images/2451/4808/95763.png)
. Substitute both values into the equation and solve for ![[{\rm OH^{-}}]](/tpl/images/2451/4808/6dddd.png)
.![\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}](/tpl/images/2451/4808/195e3.png)
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What is the OH- of {H+} = 4.0 x 10 to the power of -8...
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