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

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

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Answer 1 · 2018-04-10T08:40:57

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

Explanation:

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

$p H + p O H = p K_{w}$ $pH+pOH=pK_w$

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

$p K_{w} = - {log}_{10} [K_{w}]$ $pK_w=-log_10[K_w]$

From this, we can convert $p H$ $pH$ , the measure of $H_{3} O^{+}$ $H_3O^+$ ions, into $p O H$ $pOH$ , the measure of $O H^{-}$ $OH^-$ ions in the seawater:

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

Then we know that:
$p O H = - {log}_{10} [O H^{-}]$ $pOH=-log_10[OH^-]$
So to rearrange the equation to solve for $[O H^{-}]$ $[OH^-]$ :

$10^{- p O H} = [O H^{-}]$ $10^(-pOH)= [OH^-]$
Hence:

$10^{- 5.9} = [O H^{-}] \approx 1.26 \times 10^{- 6} m o l d m^{- 3}$ $10^(-5.9)=[OH^-] approx 1.26 times 10^-6 mol dm^-3$

Answer 2 · 2018-04-10T12:37:04

$1.26*10^-6 \ "M"$

Explanation:

Well, the $"pH"$ of the seawater is $8.1$ , and so its $"pOH"$ is:

$"pOH"=14-8.1$

$=5.9$

The $"pOH"$ of a substance is related through the equation,

$"pOH"=-log[OH^-]$

$[OH^-]$ is the hydroxide ion concentration in terms of molarity.

And so, we got:

$-log[OH^-]=5.9$

$[OH^-]=10^-5.9$

$~~1.26*10^-6 \ "M"$

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

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