To one decimal place, what is the
pOH of a 0.000259 molar aqueous HClO₄ solution at 25 degrees Celsius? Assume HClO₄ ionizes
The pOH of a solution is defined as
the negative log of the concentration of OH⁻, or hydroxide. But we’re only given the
concentration of HClO₄, which is an acid. In any aqueous solution, water can
react with itself in a process known a self-ionization or autoionization. In this equilibrium reaction, water
reacts to form H₃O⁺, or hydronium, and OH⁻, or hydroxide.
The equilibrium expression for this
reaction is the concentration of hydronium times the concentration of hydroxide
divided by the concentration of water squared. However, the concentration of water
doesn’t change significantly, so we don’t actually need to include it in this
equilibrium expression. The value of the equilibrium
constant for this reaction, which we call K w, has a value of 1.02 times 10 to the
minus 14 at 25 degrees Celsius.
If we rearrange this equation by
dividing both sides by the concentration of hydronium, we’ll have an expression for
the concentration of hydroxide, which we need to find the pOH. Now, we need to find the
concentration of hydronium. When HClO₄ is put into water, it
ionizes to form hydronium H₃O⁺ and ClO₄⁻. We’re told in the problem that
HClO₄ ionizes completely. Since HClO₄ in H₃O⁺ have a one to
one stoichiometric ratio, the concentrations of HClO₄ and H₃O⁺ must be equal.
Now, we have everything we need to
find the concentration of hydroxide. If we plug everything in, we’ll
find that the concentration of hydroxide is 3.9382 times 10 to the minus 11
molar. Now, we can find the pOH by taking
the negative log of this concentration of hydroxide. Now, if we plug everything in,
we’ll see that the pOH is 10.405. So, rounded to one decimal place,
the pOH of our HClO₄ solution is 10.4.