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

Aug 12, 2016

$\text{pH} = 7.02$

#### 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

${K}_{W} = {10}^{- 14} \to$ water's ionization constant

In pure water, water undergoes self-ionization to form hydronium cations, ${\text{H"_3"O}}^{+}$ and hydroxide anions, ${\text{OH}}^{-}$, as described by the following equilibrium reaction

$2 {\text{H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

By definition, the ionization constant is equal to

${K}_{W} = \left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right]$

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

${K}_{w} = x \cdot x = {x}^{2}$

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 $\text{ " ["OH"^(-)] = 10^(-7)"M}$

at room temperature. Now, your solution contains sodium hydroxide, $\text{NaOH}$, a strong base that dissociates completely to produce hydroxide anions in a $1 : 1$ mole ratio.

["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

${\text{ "2"H"_ 2"O"_ ((l)) rightleftharpoons" " "H"_ 3"O"_ ((aq))^(+) " "+" " " ""OH}}_{\left(a q\right)}^{-}$

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

${K}_{W} = y \cdot \left({10}^{- 8} + y\right)$

${K}_{w} = {y}^{2} + {10}^{- 8} \cdot y$

This will get you

${y}^{2} + {10}^{- 8} \cdot y - {10}^{- 14} = 0$

This quadratic equation will produce two solutions, one positive and one negative. Since $y$ represents concentration, pick the positive one

$y = 9.51 \cdot {10}^{- 8}$

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

$\text{pH} = - \log \left(9.51 \cdot {10}^{- 8}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{7.02} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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.

Aug 14, 2016

$\textsf{p H = 7.02}$

#### 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:

$\textsf{{H}_{2} O r i g h t \le f t h a r p \infty n s {H}^{+} + O {H}^{-}}$

For which $\textsf{{K}_{w} = \left[{H}^{+}\right] \left[O {H}^{-}\right] = {10}^{- 14} \textcolor{w h i t e}{x} {\text{mol"^2."l}}^{- 2}}$ at $\textsf{{25}^{\circ} C}$.

This tells us that, in pure water, the hydrogen and hydroxide ion concentrations are $\textsf{{10}^{- 7} \textcolor{w h i t e}{x} \text{mol/l}}$ respectively.

To get the total hydroxide concentration you might think you simply add $\textsf{{10}^{- 7}}$ and $\textsf{{10}^{- 8}}$. 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 $\textsf{O {H}^{-}}$ ions. The system will act to oppose that change by reducing the number of $\textsf{O {H}^{-}}$ ions and shifting to the left.

The reaction quotient $\textsf{Q}$ is given by $\textsf{\left[{H}^{+}\right] \left[O {H}^{-}\right]}$. This is now greater than $\textsf{{K}_{w}}$ so the system responds by shifting the position of equilibrium to the left such that $\textsf{Q = {K}_{w}}$.

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))

$\textsf{\textcolor{red}{C} \textcolor{w h i t e}{\times \times \times x} \text{ "-xcolor(white)(xxx)" } - x}$

$\textsf{\textcolor{red}{E} \textcolor{w h i t e}{\times \times \times \times} \text{ "(10^(-7)-x)" } \left(1.1 \times {10}^{- 7} - x\right)}$

This gives us:

$\textsf{\left({10}^{-} 7 - x\right) \left(1.1 \times {10}^{- 7} - x\right) = {10}^{- 14}}$

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

$\textsf{x = 0.49 \times {10}^{- 8}}$

We can now get the equilibrium concentration of $\textsf{\left[{H}^{+}\right]}$:

$\textsf{\left[{H}^{+}\right] = {10}^{- 7} - \left(0.49 \times {10}^{- 8}\right) = 9.51 \times {10}^{- 8} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left[9.51 \times {10}^{- 8}\right] = \textcolor{red}{7.02}}$

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