Question

In a solution at 25°C, the [H+] is 3.5 × 10−6 M. What is the [OH−]?

Answer 1

Answer on Question #42131 - Chemistry - Inorganic Chemistry

Question

In a solution at $25^{\circ}\mathrm{C}$, the $\left[\mathrm{H}^{+}\right]$ is $3.5 \times 10^{-6} \mathrm{M}$. What is the $\left[\mathrm{OH}^{-}\right]$?

Answer:

Water itself is a weak acid and a weak base. It dissociates according to the equilibrium:

$$2 \mathrm{H}_{2} \mathrm{O} \leftrightarrow \mathrm{H}_{3} \mathrm{O}^{+}_{(\mathrm{aq})} + \mathrm{OH}^{-}_{(\mathrm{aq})}$$

with a dissociation constant, $K_{\mathrm{W}}$ defined as

$$K_{\mathrm{W}} = \left[ \mathrm{H}^{+} \right] \left[ \mathrm{OH}^{-} \right],$$

where $\left[\mathrm{H}^{+}\right]$ stands for the concentration of the aquated hydronium ion and $\left[\mathrm{OH}^{-}\right]$ represents the concentration of the hydroxide ion. $K_{\mathrm{W}}$ has a value of about $10^{-14}$ at $25^{\circ}\mathrm{C}$.

Therefore the concentration of hydroxide ions in the solution equals:

$$\left[ \mathrm{OH}^{-} \right] = \frac{K_{\mathrm{W}}}{\left[ \mathrm{H}^{+} \right]} = \frac{10^{-14}}{3.5 \times 10^{-6}} = 2.86 \times 10^{-9} \mathrm{M}$$

Answer: $\left[\mathrm{OH}^{-}\right] = 2.86 \times 10^{-9} \mathrm{M}$

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