Question

What is the pH of a solution of `10^(-8)` M sodium hydroxide ?

Answer 1

`"pH" = 7.02`

Explanation:

Right from the start, you should be able to predict that the pH of the solution will be higher than `7`. This is the case because the solution contains a strong base.

You could look at the concentration of the base and say that the pH will only be slightly higher than `7`, but it must come out to be higher than `7`.

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, `"H"_3"O"^(+)` and hydroxide anions, `"OH"^(-)`, as described by the following equilibrium reaction

`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 `x` to be the equilibrium concentration of hydronium cations and hydroxide anions in pure water, you can say that you have

`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, `"NaOH"`, a strong base that dissociates completely to produce hydroxide anions in a `1:1` mole ratio.

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 `y` to be the concentrations of hydronium and hydroxide ions produced by the self-ionization reaction, you can say that you have

`" "2"H"_ 2"O"_ ((l)) rightleftharpoons" " "H"_ 3"O"_ ((aq))^(+) " "+" " " ""OH"_ ((aq))^(-)`

`color(purple)("I")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(0)aaaaaaaaaaaaacolor(black)(10^(-8))`
`color(purple)("C")color(white)(aaaaaacolor(black)(-)aaaaaaacolor(black)((+y))aaaaaaaaacolor(black)((10^(-8) + y))`
`color(purple)("E")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(y)aaaaaaaaaaacolor(black)(10^(-8) + y)`

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` represents concentration, pick the positive one

`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.

Answer 2

`sf(pH=7.02)`

Explanation:

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:

`sf(H_2OrightleftharpoonsH^++OH^-)`

For which `sf(K_w=[H^+][OH^-]=10^(-14)color(white)(x)"mol"^2."l"^(-2))` at `sf(25^@C)`.

This tells us that, in pure water, the hydrogen and hydroxide ion concentrations are `sf(10^(-7)color(white)(x)"mol/l")` respectively.

To get the total hydroxide concentration you might think you simply add `sf(10^(-7))` and `sf(10^(-8))`. This is not the case as these will not be equilibrium concentrations.

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 `sf(10^-8color(white)(x)"mol/l"`.

We have now disturbed a system at equilibrium by adding extra `sf(OH^-)` ions. The system will act to oppose that change by reducing the number of `sf(OH^-)` ions and shifting to the left.

The reaction quotient `sf(Q)` is given by `sf([H^+][OH^-])`. This is now greater than `sf(K_w)` so the system responds by shifting the position of equilibrium to the left such that `sf(Q=K_w)`.

We can set up an ICE table using equilibrium concentrations to show this:

#" "sf(H_2O" "rightleftharpoons" "H^+" "+" "OH^-) #

`sf(color(red)(I)color(white)(xxxxxxxxx)" "10^-7" "(10^(-7)+10^(-8))`

`sf(color(red)(C)color(white)(xxxxxxx)" "-xcolor(white)(xxx)" "-x)`

`sf(color(red)(E)color(white)(xxxxxxxx)" "(10^(-7)-x)" "(1.1xx10^(-7)-x))`

This gives us:

`sf((10^-7-x)(1.1xx10^(-7)-x)=10^(-14))`

If we multiply this out we get a quadratic equation so the quadratic formula can be used to solve for `sf(x)`. I won't go into that here but it gives, ignoring the -ve root:

`sf(x=0.49xx10^(-8))`

We can now get the equilibrium concentration of `sf([H^+])`:

`sf([H^+]=10^(-7)-(0.49xx10^(-8))=9.51xx10^(-8)color(white)(x)"mol/l")`

`sf(pH=-log[H^+]=-log[9.51xx10^(-8)]=color(red)(7.02))`

This is an example of "The Common Ion Effect", hydroxide being the common ion in question.