What is the [#OH^-#] in a solution of pH 3.00?

1 Answer
Apr 2, 2017

Answer:

#[OH^"-"]=10^(-11)#

Explanation:

For this answer, we use the formula:
#pH+pOH =14#
We can write that like this:
#pOH =14-pH#

Since pH=3 is given, we calculate pOH
#pOH=14-3=11#

Now the #[OH^"-"]# can be calculated from the pOH by using this formula: #[OH^"-"]=10^(-pOH)#
We obtain: #[OH^"-"]=10^(-11)#
For more information about pH, check here!

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How is #pH+pOH =14# established?
In water, the following (ionization) reaction occurs:
#2 H_"2"O -> H_"3"O^"+" + OH^"-"#

Therefore the equilibrium can be written like
#K_"c"=([H_"3"O^"+"]*[OH^"-"])/[H_"2"O]#
Since water is the solvent here, we do not take the concentration of water into consideration (is very big). We obtain the following expression:
#K_"c"=[H_"3"O^"+"]*[OH^"-"]#

The #K_"c"# in this equation represents a special number because we talk about the ionisation of water. Therefore we denote #K_"c"# as #K_"w"#. The value of the #K_"w"# is measured at 25°C.
#K_"w" (25°C) = 1*10^(-14)#
This means we can say:
#K_"c"=K_"w"=[H_"3"O^"+"]*[OH^"-"]=1*10^(-14)#

To get from the #[H_"3"O^"+"]# (concentration #H_"3"O^"+"#) to the pH, we use the following formula:
#pH=- log[H_"3"O^"+"]#
The same is true for the #[OH^"-"]#, since we define pOH as
#pOH=-log[OH^"-"]#

Now if we take the Log from both sides of the #K_"w"# equation, we get:
#log(1*10^(-14))=log([H_"3"O"]*[OH^-])#
A mathematics rule tells us that multiplying inside the logarithm function is the same as adding these logarithms. Therefore we get
#log(10^(-14))=log[H_"3"O"]+log[OH^-]#

And now we can use the definitions of pOH and OH! We get:
#log(10^(-14))=-pH -pOH#
with #log(10^(-14))=-14# we get our function
#-pH-pOH =-14#
Which is the same as
#pH+pOH=14#