# Question 1108e

Feb 20, 2016

Yes, the pH is equal to $6$.

#### Explanation:

This is a perfect example of a problem that wants to test your understanding of how the concentrations of hydronium and hydroxide ions determine a solution's pH.

When a problem gives a different value for water ion product, ${K}_{W}$, your first impulse should be to figure out the concentrations of hydroxide and hydronium ions present in a neutral sample of pure water at that given temperature.

As you know, the ion product of water is equal to

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

At ${80}^{\circ} \text{C}$, you know that this will be equal to

${K}_{W} = \left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right] = {10}^{- 12}$

SIDE NOTE I'll skip the units for the ion product constant, they are not important in this context

But you should also know that a neutral solution will always have equal concentrations of hydronium and hydroxide anions, which means that

${K}_{w} = x \cdot x = {x}^{2} = {10}^{- 12}$

x = ["H"_3"O"^(+)] = ["OH"^(-)] = sqrt(10^(-12)) = 10^(-6)"M"

Now take a look at the molarity of that sodium hydroxide solution. The concentration of hydroxide anions in a ${\text{1 dm}}^{3}$ sample of that is so small that you can actually consider it negligible.

Why is that so?

Well, you know that at ${80}^{\circ} \text{C}$, ${\text{1 dm}}^{3}$ of pure water contains ${10}^{- 6}$ moles of hydroxide anions. Remember, when you're working with a ${\text{1-dm}}^{3}$ sample, molarity and number of moles are interchangeable.

In order to have a ${10}^{- 9} \text{M}$ solution of sodium hydroxide, you need to add ${10}^{- 9}$ moles of sodium hydroxide in 1 dm"^3 of pure water.

But ${10}^{- 9}$ moles of hydroxide anions is a very small amount compared with ${10}^{- 6}$ moles of hydroxide anions that are already present in the pure water.

In other words, you will have

n_(OH^(-)) = overbrace(10^(-6)"moles OH"^(-))^(color(purple)("present in pure water")) + overbrace(10^(-9)"moles OH"^(-))^(color(purple)("added to the water"))

${n}_{O {H}^{-}} \approx {10}^{- 6} \text{moles}$

In a ${\text{1-dm}}^{3}$ sample, this will give you

["OH"^(-)] ~~ 10^(-6)"M"

Of course, this translates to

$\text{pOH} = - \log \left({10}^{- 6}\right) = 6$

and

$\text{pH} = 12 - 6 = \textcolor{g r e e n}{6}$

Remember, the solution is neutral at $\text{pH} = 6$ at ${80}^{\circ} \text{C}$, since you have

["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-6)"M"#

As a final note, chemistry is not about calculations. Focus on understanding what's going on first, then worry about integrals, logs, equations, formulas, derivatives, and things of that nature.

If you focus on calculations too much, you'll end up not seeing the forest for the trees.