# Consider the titration of 100 ml of 0.1 M HCl with 0.1 M NaOH. Calculate the pH of the solution at the following stages: ?

## At the beginning, when no $N a O H$ is added When $50$ $m l$ of $0.1$ $M$ is added When $98$ $m l$ of $0.1$ $M$ is added When $99.9$ $m l$ of $0.1$ $M$ is added When $100$ $m l$ of $0.1$ $M$ is added When $100.1$ $m l$ of $0.1$ $M$ is added

Mar 22, 2016

Here's what I got.

#### Explanation:

You're titrating hydrochloric acid, $\text{HCl}$, a strong acid, with sodium hydroxide, $\text{NaOH}$, a strong base, so right from the start you should know that the pH at equivalence point must be equal to $7$.

Hydrochloric acid and sodium hydroxide react in a $1 : 1$ mole ratio to form water and aqueous sodium chloride

${\text{HCl"_text((aq]) + "NOH"_text((aq]) -> "NaCl"_text((aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

The net ionic equation for this reaction looks like this

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

Now, the equivalence point corresponds to a complete neutralization, i.e. when you add enough strong base to completely consume all the acid present in the solution.

The $1 : 1$ mole ratio tells you that at the equivalence point, the solution must contain equal numbers of moles of strong acid and strong base.

Your tool of choice here will the equation

color(blue)(|bar(ul(color(white)(a/a)"pH" = -log(["H"_3"O"^(+)])color(white)(a/a)|)))

Now, it's very important to realize that the volume of the solution will increase as you add the strong base solution. Keep this in mind when calculating the molarity of the hydronium ions.

So, let's start calculating the corresponding pH

$1. \textcolor{p u r p \le}{\underline{\text{Before NaOH is added}}}$

Since hydrochloric acid is a strong acid, it dissociates completely in aqueous solution to form hydronium cations and chloride anions. More specifically, you have

["H"_3"O"^(+)] = ["HCl"] = "0.1 M"

This means that the pH of the solution before any strong base is added will be equal to

$\text{pH} = - \log \left(0.1\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 1 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$2. \textcolor{p u r p \le}{\underline{\text{After 50 mL of NaOH are added}}}$

So, use the definition of molarity to determine how many moles of hydronium ions, i.e. hydrochloric acid, you have in the initial solution

$\textcolor{b r o w n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"soluteion" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

${n}_{{H}_{3} {O}^{+}} = {\text{0.1 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.010 moles H"_3"O}}^{+}$

Calculate how many moles of hydroxide anions, i.e. sodium hydroxide, are delivered to the solution by adding $\text{50 mL}$ of $\text{0.1-M}$ sodium hydroxide solution

${n}_{O {H}^{-}} = {\text{0.1 mol" color(red)(cancel(color(black)("L"^(-1)))) * 50 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.0050 moles OH}}^{-}$

The hydroxide anions will be completely consumed by the reaction, leaving behind

${n}_{O {H}^{-}} = 0 \to$ completely consumed

${n}_{{H}_{3} {O}^{+}} = {\text{0.010 moles" - "0.0050 moles" = "0.0050 moles H"_3"O}}^{+}$

The total volume of the solution will be

${V}_{\text{total" = "100 mL" + "50 mL" = "150 mL}}$

The concentration of hydronium ions will be

["H"_3"O"^(+)] = "0.0050 moles"/(150 * 10^(-3)"L") = "0.0333 M"

The pH of the solution will be

$\text{pH} = - \log \left(0.0333\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 1.48 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$3. \textcolor{p u r p \le}{\underline{\text{After 98 mL of NaOH are added}}}$

This corresponds to adding an additional $\text{48 mL}$ of strong base, which is equivalent to

${n}_{O {H}^{-}} = {\text{0.1 mol" color(red)(cancel(color(black)("L"^(-1)))) * 48 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.0048 moles OH}}^{-}$

Once again, the hydroxide anions will be completely consumed by the reaction, leaving you with

${n}_{O {H}^{-}} = 0 \to$ completely consumed

${n}_{{H}_{3} {O}^{+}} = 0.0050 - 0.0048 = {\text{0.00020 moles H"_3"O}}^{+}$

The total volume of the solution will be

${V}_{\text{total" = "150 mL" + "48 mL" = "198 mL}}$

The concentration of hydronium ions will be

["H"_3"O"^(+)] = "0.00020 moles"/(198 * 10^(-3)"L") = "0.001010 M"

The pH of the solution will be

$\text{pH} = - \log \left(0.001010\right) \approx \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 3 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$4. \textcolor{p u r p \le}{\underline{\text{After 99.9 mL of NaOH are added}}}$

This corresponds to adding an additional $\text{1.9 mL}$ of strong base, which is equivalent to

${n}_{O {H}^{-}} = {\text{0.1 mol" color(red)(cancel(color(black)("L"^(-1)))) * 1.9 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.00019 moles OH}}^{-}$

The number of moles of hydroxide anions is still smaller than the number of moles of hydronium cations, so the hydroxide anions will once again by completely consumed here.

The resulting solution will contain

${n}_{O {H}^{-}} = 0 \to$ completely consumed

${n}_{{H}_{3} {O}^{+}} = 0.00020 - 0.00019 = {\text{0.000010 moles H"_3"O}}^{+}$

The *total volume of the solution will be

${V}_{\text{total" = "198 mL" + "1.9 mL" = "199.9 mL}}$

The concentration of hydronium cations will be

["H"_3"O"^(+)] = "0.000010 moles"/(199.9 * 10^(-3)"L") = "0.00005 M"

The pH of the solution will be

$\text{pH} = - \log \left(0.00005\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 4.3 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

5. color(purple)(ul("After 100 mL of NaOH are added")) -> "EQUIVALENCE POINT"

Adding an additional $\text{0.1 mL}$ of strong base to the solution will correspond to a complete neutralization.

All the moles of hydronium cations that were initially present in the solution are now completely consumed. The resulting solution is neutral, since it only contains water and aqueous sodium chloride.

The concentration of hydronium cations corresponds to that of pure water at room temperature, $1.0 \cdot {10}^{- 7} \text{M}$. The pH of the solution is

$\text{pH} = - \log \left(1.0 \cdot {10}^{- 7}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 7 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$6. \textcolor{p u r p \le}{\underline{\text{After 100.1 mL of NaOH are added}}}$

This corresponds to adding an additional $\text{0.1 mL}$ of strong base to a neutral solution, so you can expect the pH to be higher than $7$.

You will have

${n}_{O {H}^{-}} = {\text{0.1 mol" color(red)(cancel(color(black)("L"^(-1)))) * 0.1 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.00001 moles OH}}^{-}$

The total volume of the solution will be

${V}_{\text{total" = "200 mL" + "0.1 mL" = "201 mL}}$

The concentration of hydroxide anions will be

["OH"^(-)] = "0.000010 moles"/(200.1 * 10^(-3)"L") = "0.000049975 M"

The pOH of the solution will be

$\text{pOH} = - \log \left(0.000049975\right) \approx 4.3$

Since you know that, at room temperature

$\textcolor{b r o w n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH " + " pOH} = 14} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

you can say that the pH of the solution will be

$\text{pH} = 14 - 4.3 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 9.7 \textcolor{w h i t e}{\frac{a}{a}} |}}}$