# Question 1fbd7

##### 1 Answer
Dec 29, 2017

(a) $\text{pH} = 13.65$

(b) $\text{pH} = 7$

#### Explanation:

The idea here is that you're dealing with neutralization reactions that may or may not be complete, depending on the concentrations of hydronium cations and hydroxide anions that you mix.

If the neutralization reaction is complete, then the $\text{pH}$ of the resulting solution will be equal to $7$ at ${25}^{\circ} \text{C}$. If the neutralization is not complete, then the concentration of hydronium cations present after you mix the two solutions will give you the $\text{pH}$ of the resulting solution.

As you know, when a neutralization reaction takes place, you have

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

For the first reaction, you have one solution that has

$\text{pH" = 1 implies ["H"_3"O"^(+)] = 1 * 10^(-1)color(white)(.)"M}$

Similarly, the second solution has

$\text{pH" = 14 implies "pOH} = 14 - 14 = 0$

This gets you

["OH"^(-)] = 1 * 10^0color(white)(.)"M"

When you mix equal volumes of these two solutions, the hydronium cations will act as the limiting reagent, i.e. they will be completely consumed by the neutralization reaction.

This happens because you're not mixing equal concentrations of hydronium cations and hydroxide anions, so the neutralization will not be complete, i.e. the $\text{pH}$ of the resulting solution will be $> 7$ because the resulting solution will contain excess hydroxide anions.

After the reaction takes place, you will have

["OH"^(-)] = 1 * 10^0color(white)(.)"M" - 1 * 10^(-1)color(white)(.)"M"

["OH"^(-)] = 9 * 10^(-1)color(white)(.)"M"

Don't forget that when you mix equal volumes of the two solutions, the total volume of the resulting solution will be twice that of each solution.

This implies that the concentration of the hydroxide anions will be halved

["OH"^(-)] = (9 * 10^(-1)color(white)(.)"M")/2 = 4.5 * 10^(-1)color(white)(.)"M"

Use the concentration of the hydroxide anions to find the $\text{pH}$ of the solution

"pH" = 14 - [-log(["OH"^(-)])]

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH}}}} = 14 + \log \left(4.5 \cdot {10}^{- 1}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{13.65}}}$

$\textcolor{w h i t e}{\frac{a}{a}}$
For the second reaction, you have one solution that has

$\text{pH" = 5 implies ["H"_3"O"^(+)] = 1 * 10^(-5)color(white)(.)"M}$

The second solution has

$\text{pH" = 9 implies "pOH} = 14 - 9 = 5$

This gets you

["OH"^(-)] = 1 * 10^(-5)color(white)(.)"M"

This time, you're mixing equal volumes of a strong acid and of a strong base that contain equal concentrations of hydronium cations and hydroxide anions, respectively.

This tells you that the neutralization reaction will be complete, meaning that the resulting solution will not contain any excess hydronium cations or hydroxide anions.

All the hydronium cations delivered by the strong acid and all the hydroxide anions delivered by the strong base will neutralize each other to produce water.

Consequently, you can say that at ${25}^{\circ} \text{C}$, the resulting solution will have

["H"_3"O"^(+)] = ["OH"^(-)] = 1 * 10^(-7)color(white)(.)"M"#

These concentrations of hydronium cations and hydroxide anions are produced by the auto-ionization of water.

The resulting solution, which is neutral on account of the fact that it contains equal concentrations of hydronium cations and hydroxide anions, will have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = 7}}}$

The volume of the resulting solution is not important here because it doesn't contain any excess hydronium cations or hydroxide anions that you need to keep track of.