# Question 71616

Nov 18, 2015

Here's what I got.

#### Explanation:

So, you're titrating lithium hydroxide, a strong base, with hydrochloric acid, a strong acid.

Right from the start, you can say that the pH of the solution after adding $\text{35 mL}$ of strong acid will be equal to $7$. That happens because equal amounts of strong base and strong acid will lead to a complete neutralization.

So, let's break this titration down step by step.

• Part A - before adding $\text{HCl}$

Since lithium hydroxide is a strong base, it will dissociate completely in aqueous solution to produce lithium cations, ${\text{Li}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$.

${\text{LiOH"_text((aq]) -> "Li"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

Since you get one mole of hydroxide anions for every one mole of lithium hydroxide, it follows that the concentration of hydroxide anions will match that of lithium hydroxide

["OH"^(-)] = ["LiOH"] = "0.110 M"

The $\text{pOH}$ of the solution will be

"pOH" = - log(["OH"^(-)])

$\text{pOH} = - \log \left(0.110\right) = 0.96$

The pH of the solution will be

$\text{pH" = 14 - "pOH} = 14 - 0.96 = \textcolor{g r e e n}{13.04}$

• Part B - after adding $\text{13.5 mL}$ of $\text{HCl}$

Now, before moving forward, it's important to understand what happens when a strong base reacts with a strong acid. In this case, the net ionic equation for this reaction will be

${\text{OH"_text((aq])^(-) + "H"_3"O"_text((aq])^(+) -> 2"H"_2"O}}_{\textrm{\left(l\right]}}$

One mole of hydroxide anions will be neutralized by one mole of hydronium cations, resulting in the formation of water.

Use the molarity and volume of the lithium hydroxide solution to determine how many moles of hydroxide anions you have in the initial solution

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{O {H}^{-}} = {\text{0.110 M" * 35 * 10^(-3)"L" = "0.00385 moles OH}}^{-}$

Do the same for the added hydrochloric acid solution - keep in mind that one mole of hydrochloric acid produces one mole of hydronium ions.

${n}_{{H}_{3} {O}^{+}} = {\text{0.110 M" * 13.5 * 10^(-3)"L" = "0.001485 moles H"_3"O}}^{+}$

So, what will happen when these two solutions are mixed? The species that has fewer number of moles will be **completely consumed by the neutralization reaction.

This implies that the number of moles of the species that was in excess will decrease by the same number as you had number of moles of the consumed species.

In this case, the hydronium ions will be completely consumed and you'll be left with

${n}_{{H}_{3} {O}^{+}} = \text{0 moles}$

${n}_{O {H}^{-}} = 0.00385 - 0.001485 = {\text{0.002365 moles OH}}^{-}$

The volume of the solution will be

${V}_{\text{sol" = "35 mL" + "13.5 mL" = "48.5 mL}}$

The concentration of hydroxide ions will now be

["OH"^(-)] = "0.002365 moles"/(48.5 * 10^(-3)"L") = "0.04845 M"

The pH of the solution will be

$\text{pH} = 14 - \left(- \log \left(0.04845\right)\right) = \textcolor{g r e e n}{12.69}$

• Part C - after adding $\text{25.5 mL}$ of $\text{HCl}$

The exact same strategy applies. Your solution now contains an extra $\text{12 mL}$ of acid, which is equivalent to

${n}_{{H}_{3} {O}^{+}} = {\text{0.110 M" * 12 * 10^(-3)"L" = "0.00132 moles H"_3"O}}^{+}$

This time, the reaction will leave you with

${n}_{{H}_{3} {O}^{+}} = \text{0 moles}$

${n}_{O {H}^{-}} = 0.002365 - 0.00132 = {\text{0.001405 moles OH}}^{-}$

The volume of the solution will now be

${V}_{\text{sol" = "48.5 mL" + "12 mL" = "60.5 mL}}$

The concentration of hydroxide ions will be

["OH"^(-)] = "0.001405 moles"/(60.5 * 10^(-3)"L") = "0.02322 M"

The pH of the solution will now be

$\text{pH} = 14 - \left(- \log \left(0.02322\right)\right) = \textcolor{g r e e n}{12.37}$

• Part D - after adding $\text{35 mL}$ of $\text{HCl}$

This time, you're adding an extra $\text{9.5 mL}$ of acid. This
is equivalent to adding $0.001405$ of hydronium ions - calculate it to make sure that it's correct!

Equal number of moles of hydroxide and hydronium ions will result in a neutral solution.

$\text{pH} = \textcolor{g r e e n}{7}$

• Part E - after adding $\text{43.5 mL}$ of $\text{HCl}$

At this point, the base is completely consumed. Adding acid will result in an acidic solution. You're adding an extra $\text{8.5 mL}$ of acid, which is equivalent to

${n}_{{H}_{3} {O}^{+}} = {\text{0.110 M" * 8.5 * 10^(-3)"L" = "0.000935 moles H"_3"O}}^{+}$

The total volume of the solution is

${V}_{\text{total" = "70 mL" + "8.5 mL" = "78.5 mL}}$

The concentration of hydronium ions is

["H"_3"O"^(+)] = "0.000935 moles"/(78.5 * 10^(-3)"L") = "0.01191 M"

The pH of the solution will be

"pH" = -log(["H"_3"O"^(+)])#

$\text{pH} = - \log \left(0.01191\right) = \textcolor{g r e e n}{1.94}$

• Part D - after adding $\text{50 mL}$ of $\text{HCl}$

This one I'll leave to you as practice.