Seawater has a pH of 8.100. What is the concentration of OH–?

Apr 10, 2018

${10}^{-} 5.9 \approx 1.26 \times {10}^{-} 6 m o l {\mathrm{dm}}^{-} 3$

Explanation:

If the $p H$ is 8.1, and we assume that this is measured under standard conditions, we can use the relationship:

$p H + p O H = p {K}_{w}$

At 25 degrees celcius, $p {K}_{w} = 14$
Where ${K}_{w}$ is the dissociation constant for water- $1.0 \times {10}^{-} 14$,(At 25 degrees C) but $p {K}_{w}$ is the negative logarithm of ${K}_{w}$.

$p {K}_{w} = - {\log}_{10} \left[{K}_{w}\right]$

From this, we can convert $p H$, the measure of ${H}_{3} {O}^{+}$ ions, into $p O H$, the measure of $O {H}^{-}$ ions in the seawater:

$p H + p O H = 14$
$8.1 + p O H = 14$
$p O H = 5.9$

Then we know that:
$p O H = - {\log}_{10} \left[O {H}^{-}\right]$
So to rearrange the equation to solve for $\left[O {H}^{-}\right]$:

${10}^{- p O H} = \left[O {H}^{-}\right]$
Hence:

${10}^{- 5.9} = \left[O {H}^{-}\right] \approx 1.26 \times {10}^{-} 6 m o l {\mathrm{dm}}^{-} 3$

Apr 10, 2018

$1.26 \cdot {10}^{-} 6 \setminus \text{M}$

Explanation:

Well, the $\text{pH}$ of the seawater is $8.1$, and so its $\text{pOH}$ is:

$\text{pOH} = 14 - 8.1$

$= 5.9$

The $\text{pOH}$ of a substance is related through the equation,

$\text{pOH} = - \log \left[O {H}^{-}\right]$

• $\left[O {H}^{-}\right]$ is the hydroxide ion concentration in terms of molarity.

And so, we got:

$- \log \left[O {H}^{-}\right] = 5.9$

$\left[O {H}^{-}\right] = {10}^{-} 5.9$

$\approx 1.26 \cdot {10}^{-} 6 \setminus \text{M}$