What is the pH of a solution of #10^(-8)# M sodium hydroxide ?
Right from the start, you should be able to predict that the pH of the solution will be higher than
You could look at the concentration of the base and say that the pH will only be slightly higher than
The key here is the self-ionization of water, which at room temperature has an equilibrium constant equal to
#K_W = 10^(-14) ->#water's ionization constant
In pure water, water undergoes self-ionization to form hydronium cations,
#2"H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-)#
By definition, the ionization constant is equal to
#K_W = ["H"_3"O"^(+)] * ["OH"^(-)]#
If you take
#K_w = x * x = x^2#
This gets you
#x = sqrt(K_W) = sqrt(10^(-14)) = 10^(-7)#
So, the self-ionization of water produces
#["H"_3"O"^(+)] = 10^(-7)"M "#and #" " ["OH"^(-)] = 10^(-7)"M"#
at room temperature. Now, your solution contains sodium hydroxide,
Therefore, you're adding
#["OH"^(-)] = ["NaOH"] = 10^(-8)"M"#
to pure water. The key now is the fact that the self-ionization of water will still take place, but because the concentration of hydroxide anions has increased, the equilibrium will produce fewer moles of hydronium and hydroxide ions.
If you take
#" "2"H"_ 2"O"_ ((l)) rightleftharpoons" " "H"_ 3"O"_ ((aq))^(+) " "+" " " ""OH"_ ((aq))^(-)#
This time, the ionization constant will be equal to
#K_W = y * (10^(-8) + y)#
#K_w = y^2 + 10^(-8) * y#
This will get you
#y^2 + 10^(-8) * y - 10^(-14) = 0#
This quadratic equation will produce two solutions, one positive and one negative. Since
#y = 9.51 * 10^(-8)#
This means that the equilibrium concentration of hydronium cations will be
#["H"_3"O"^(+)] = 9.51 * 10^(-8)"M"#
The pH of the solution is given by
#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#
Plug in your value to find
#"pH" = - log(9.51 * 10^(-8)) = color(green)(|bar(ul(color(white)(a/a)color(black)(7.02)color(white)(a/a)|)))#
As you can see, the pH of the solution is consistent with the fact that your solution contains a strong base, albeit in a very small concentration.
This is a tiny concentration for NaOH so I would expect the pH to be close to 7 or slightly greater.
Because the concentration is so small, we must take into account the ions that are produced from the auto - ionisation of water:
This tells us that, in pure water, the hydrogen and hydroxide ion concentrations are
To get the total hydroxide concentration you might think you simply add
We need to apply Le Chatelier's Principle.
If we do a thought experiment we can imagine some pure water to which a tiny amount of sodium hydroxide is added such that the concentration of NaOH is
We have now disturbed a system at equilibrium by adding extra
The reaction quotient
We can set up an ICE table using equilibrium concentrations to show this:
This gives us:
If we multiply this out we get a quadratic equation so the quadratic formula can be used to solve for
We can now get the equilibrium concentration of
This is an example of "The Common Ion Effect", hydroxide being the common ion in question.