Question #44429

1 Answer
Ken C.
Nov 7, 2016

#[OH^-]=1.74xx10^-6M#

Explanation:

First, we use this relationship between pH and pOH:
#"pH"+"pOH"=14#

We know that #"pH"=8.24#, so
#"pOH"=14-"pH"=14-8.24=5.76#

To determine the #[OH^-]#, we use the following relationship between #"pOH"# and #[OH^-]#:
#-log[OH^-]="pOH"#

Where "log" is the base ten logarithm. If you're not familiar with logs, they are essentially the opposite of exponents. For example, we know that #10^2=100#; that means #log100=2#. In general, #logx=y# if #10^y=x#. If you want to learn more about logs, you can check this link.

Back to the problem. We know that #"pOH"=5.76#, so using the formula above, we have:
#-log[OH^-]=5.76#

Dividing by #-1# to transfer the negative sign, we have:
#log[OH^-]=-5.76#

Finally, we raise both sides to the power of ten to cancel the logarithm, leaving us with:
#[OH^-]=10^-5.76=1.74xx10^-6M#
as our concentration.

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