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

2 Answers
Apr 10, 2018

#10^-5.9 approx 1.26 times 10^-6 mol dm^-3#

Explanation:

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

#pH+pOH=pK_w#

At 25 degrees celcius, #pK_w=14#
Where #K_w# is the dissociation constant for water- #1.0 times 10^-14#,(At 25 degrees C) but #pK_w# is the negative logarithm of #K_w#.

#pK_w=-log_10[K_w]#

From this, we can convert #pH#, the measure of #H_3O^+# ions, into #pOH#, the measure of #OH^-# ions in the seawater:

#pH+pOH=14#
#8.1+ pOH=14#
#pOH=5.9#

Then we know that:
#pOH=-log_10[OH^-]#
So to rearrange the equation to solve for #[OH^-]#:

#10^(-pOH)= [OH^-]#
Hence:

#10^(-5.9)=[OH^-] approx 1.26 times 10^-6 mol dm^-3#

Apr 10, 2018

#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"#