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

##### 2 Answers

#### Explanation:

Right from the start, you should be able to predict that the pH of the solution will be **higher than** **strong base**.

You could look at the concentration of the base and say that the pH will only be *slightly* higher than **must** come out to be 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'sionization 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 **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, **strong base** that dissociates completely to produce hydroxide anions in a

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 **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))^(-)#

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 **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.

#### 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:

For which

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 **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

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.