At 25 degrees Celsius, the
hydronium ion concentration of vinegar is approximately 4.0 times 10 to the minus
three molar. To one decimal place, what is the
corresponding value of the POH?
This problem wants us to find the
POH of vinegar, which is defined as the negative log of the concentration of
hydroxide. But the problem only gives us the
concentration of hydronium. To solve this problem, we’re going
to make use of the fact that, in any aqueous solution, water can react with itself
in a process known as self-ionization or autoionization.
In this equilibrium reaction, water
reacts to form the hydronium ion and the hydroxide ion. The equilibrium expression for this
reaction is equal to the concentration of hydronium times the concentration of
hydroxide divided by the concentration of water squared. Since the concentration of water
doesn’t significantly change, we don’t have to include it in our equilibrium
expression. The equilibrium constant for this
reaction, which we call Kw, has a value of 1.0 times 10 to the minus 14 at 25
Since the problem gives us the
concentration of hydronium, we have everything we need to solve for the
concentration of hydroxide. The first thing I’ve done here is
just switched the sides of the equation. So ultimately, we’ll end up with
the concentration of hydroxide on the left.
Now we can divide both sides by the
concentration of the hydronium ion to obtain an expression for hydroxide. Plugging in both the value of Kw
and the concentration of hydronium, we’ll find that the concentration of hydroxide
is 2.50 times 10 to the minus 12 molar. Now we can find the POH by taking
the negative log of this concentration that we just found. Plugging that in, we’ll find that
the POH is 11.602. So rounded to one decimal place,
the POH of vinegar is 11.6.