# Neutralisations and Dilutions

A LevelAQA

## Neutralisations and Dilutions

In both neutralisation and dilution, we deal with changes in the number of moles present and in the volumes of solution involved. These changes will lead to new concentrations of  chemicals, and hence effect the $\text{pH}$ of the solution. For calculations involving neutralisations and dilutions, the moles contained in solutions and the new volume must be taken into account to calculate the $\text{pH}$.

## Neutralisation: Strong Acid and Strong Base

When strong acids and bases are reacted together, the resulting mixture will have a new $\text{pH}$ depending on the concentration of $\text{OH}^-$ ions of $\text{H}^+$ ions in the solution.

To calculate the new $\text{pH}$, we must determine which of the ions is in excess and by how much.

1. Calculate the moles of both the acid and the alkali.
2. Determine which of the two has moles in excess (more moles). If the $\text{OH}^-$ ions are in excess if there are more moles of the alkali. If the acid has more moles, the $\text{H}^+$ ions are in excess.
3. Work out the difference between the two mole values calculated.
4. Work out the concentration of the ions in excess by using the equation $Concentration = \frac{Moles}{Volume} \text{ } (C=\frac{n}{V})$.
5. If the acid is in excess, use the concentration calculated to work out the $\text{pH}$. If the alkali is in excess, put the value calculated into the $\text{K}_w$ equation to find $\left[\text{H}^+\right]$ and then put that into the $\text{pH}$ equation.

Some acids and bases are what are known as diprotic. Diprotic acids are acids that can donate two hydrogen ions into solution. Sulfuric acid $(\text{H}_2\text{SO}_4)$ is a common example of a diprotic acid. Diprotic bases are bases that can accept two hydrogen ions. Calcium hydroxide $(\text{Ca(OH}_2\text{)}$ is a common example of a diprotic base.

When calculating the $\text{pH}$ of a diprotic acid or base, we need to account for the fact that $2\text{ H}^+$ ions are involved. To do this, when we calculate the moles, we multiply it by $2$.

A LevelAQA

## Neutralisation: Weak Acid and Strong Base

Just like when calculating the pH for strong acid and base neutralisations, we have to work out the excess moles to find the $\text{pH}$ of the new solution.

However, when a neutralisation involves a weak acid and strong base, we have to calculate the concentration of the anions $\left(\text{A}^-\right)$ within the solution to input it into the $\text{K}_a$ equation. We find the concentration of the anion by dividing the moles of $\text{OH}^-$ added by the new total volume:

$\left[\text{A}^-\right]=\frac{\left[\text{OH}^-\right]}{\text{Final Volume}}$

A LevelAQA

## Half Equivalence

Sometimes, you may be asked to find the $\text{pH}$ of a weak acid that has reacted with half of the volume of an alkali required for neutralisation. This is known as the half equivalence point.

To calculate the half equivalence point, we first calculate the value of $\text{K}_a$. At the half equivalence point it can be assumed that $\left[\text{HA}\right] =\left[\text{A}^-\right]$. The value of $\text{K}_a$ therefore becomes equal to that of $\left[\text{H}^+\right]$.

\begin{aligned}\text{K}_a&=\frac{\left[\text{H}^+\right]\left[\text{A}^-\right]}{\left[\text{HA}\right]}\\\text{}\\\text{K}_a&=\frac{\left[\text{H}^+\right]\left[\text{A}^-\right]}{\left[\text{A}^-\right]}\\\text{}\\\text{K}_a&=\left[\text{H}^+\right]\end{aligned}

As a result, the value of the $\text{pK}_a$ is equals $\text{pH}$ at the half equivalence point. The half equivalence point can be determined from calculation or by using a $\text{pH}$ curve.

A LevelAQA

## $\text{pH}$ of Diluted Solutions

When calculating dilutions, the new volume must be accounted for. A new volume will mean a new concentration. Changes in concentration are related by the equation:

$\text{C}_1\text{V}_1=\text{C}_2\text{V}_2$

The following equation can be used to calculate the $\text{pH}$of a diluted acid.

$\left[\text{H}^+\right]_{final}=\left[\text{H}^+\right]_{original} \times \frac{\text{Original Volume}}{\text{Final Volume}}$

To calculate the$\text{pH}$ of a diluted alkali, the following equation can be used.

$\left[\text{OH}^-\right]_{final}=\left[\text{OH}^-\right]_{original} \times \frac{\text{Original Volume}}{\text{Final Volume}}$

## Example 1: Reacting Strong Bases and Strong Acids.

$\textcolor{#00bfa8}{35\text{ cm}^3}$ of $\textcolor{#f21cc2}{0.7\text{ mol dm}^{-3}}$ of $\text{HCl}$ reacts with $\textcolor{#a233ff}{40\text{ cm}^{-3}}$ of $\textcolor{#eb6517}{0.4\text{ mol dm}^{-3}}$ of $\text{NaOH}$. Calculate the $\text{pH}$ of the resulting mixture.

[5 marks]

$\text{Moles HCl}=\textcolor{#f21cc2}{0.7}\times\frac{\textcolor{#00bfa8}{35}}{1000}=\textcolor{#008d65}{0.0245\text{ mol}}$

$\text{Moles NaOH}=\textcolor{#eb6517}{0.4}\times\frac{\textcolor{#a233ff}{40}}{1000}=\textcolor{#008d65}{0.0160\text{ mol}}$

$\text{Excess HCl}=0.0245=0.016=\textcolor{#008d65}{8.50\times10^{-3}\text{ mol}}$

$\left[\text{H}^+\right]=\frac{8.50\times10^{-3}}{\textcolor{#a233ff}{0.040}+\textcolor{#00bfa8}{0.035}}=\textcolor{#008d65}{0.113}$

$\text{pH}=-\text{log }0.113=\textcolor{#008d65}{0.946}$

A LevelAQA

## Example 2: Reactions with Diprotic Acids

$\textcolor{#00bfa8}{25\text{ cm}^3}$ of $\textcolor{#f21cc2}{0.4\text{ mol dm}^{-3}}\text{ HCl}$ is reacted with $\textcolor{#a233ff}{30\text{ cm}^3}$ of $\textcolor{#eb6517}{0.5\text{ mol dm}^{-3}}\text{ Ba(OH)}_2$. Calculate the pH of the resulting mixture.

[6 marks]

$\text{Moles HCl}=\textcolor{#f21cc2}{0.4}\times\frac{\textcolor{#00bfa8}{25}}{1000}=\textcolor{#008d65}{0.010\text{ mol}}$

$\text{Moles Ba(OH)}_2=\textcolor{#eb6517}{0.5}\times\frac{\textcolor{#a233ff}{30}}{1000}=\textcolor{#008d65}{0.015\text{ mol}}$

$\text{Moles OH}^-=0.015\times2=\textcolor{#008d65}{0.030\text{ mol}}$

$\text{Excess OH}^-=0.03-0.010=\textcolor{#008d65}{0.020\text{ mol}}$

$\left[\text{OH}^-\right]=\frac{0.020}{0.055}=\textcolor{#008d65}{0.364\text{ mol dm}^{-3}}$

\begin{aligned}\left[\text{H}^+\right]&=\frac{\text{K}_w}{\left[\text{OH}^-\right]}=\frac{10^{-14}}{0.364}\\ &=\textcolor{#008d65}{2.75\times10^{-14}}\end{aligned}

$\text{pH}=-\text{log }2.75\times10^{-14}=\textcolor{#008d65}{13.6}$

A LevelAQA

## Example 3: Reactions with a Weak Acid

$\textcolor{#00bfa8}{60\text{ cm}^3}$ of $\textcolor{#f21cc2}{0.65\text{ mol dm}^{-3}}$ of ethanoic acid is reacted with $\textcolor{#a233ff}{35\text{ cm}^3}$ of $\textcolor{#eb6517}{0.45\text{ mol dm}^{-3}}$ of $\text{NaOH}$. Calculate the $\text{pH}$ of the resulting mixture. The $\text{K}_a$ for ethanoic acid is $1.75\times10^{-5}$.

[8 marks]

$\text{Moles Ethanoic Acid}=\textcolor{#f21cc2}{0.65}\times\frac{\textcolor{#00bfa8}{60}}{1000}=\textcolor{#008d65}{0.039\text{ mol}}$

$\text{Moles NaOH} =\text{Moles OH}^-=\textcolor{#eb6517}{0.45}\times\frac{\textcolor{#a233ff}{35}}{1000}=\textcolor{#008d65}{0.016\text{ mol}}$

$\text{Excess Ethanoic Acid}=0.039-0.016=\textcolor{#008d65}{0.023\text{ mol}}$

$\left[\text{Ethanoic Acid}\right]=\frac{0.023}{0.095}=\textcolor{#008d65}{0.245\text{ mol dm}^{-3}}$

$\left[\text{Ethanoate Ions}\right]=\frac{0.016}{0.095}=\textcolor{#008d65}{0.167\text{ mol dm}^{-3}}$

\begin{aligned}\text{K}_a&=\frac{\left[\text{H}+\right]\left[\text{CH}_3\text{COO}^-\right]}{\left[\text{CH}_3\text{COOH}\right]}\\ \text{}\\\left[\text{H}^+\right]&=\frac{\text{K}_a\times\left[\text{CH}_3\text{COOH}\right]}{\left[\text{CH}_3\text{COO}^-\right]}\\ \text{}\\ &=\frac{1.75\times10^{-5}\times0.245}{0.166}\\ \text{}\\&=\textcolor{#008d65}{2.58\times10^{-5}\text{ mol dm}^{-3}}\end{aligned}

$\text{pH}=-\text{log }2.58\times10^{-5}=\textcolor{#008d65}{4.6}$

A LevelAQA

## Example 4: Calculations with the Half Equivalence Point

Calculate the $\text{pH}$ of the resulting solution when $\textcolor{#00bfa8}{30\text{ cm}^3}$ of $\textcolor{#f21cc2}{0.2\text{ mol dm}^{-3}}$ sodium hydroxide is added to $\textcolor{#a233ff}{60\text{ cm}^3}$ of $\textcolor{#eb6517}{0.2\text{ mol dm}^{-3}}$ of ethanoic acid. The Ka of ethanoic acid is $\textcolor{#0040ff}{1.7\times10^{-5}\text{ mol dm}^{-3}}$.

[2 marks]

We can deduce that the reaction is at half equivalence point from the concentrations and volumes of the reactants. Both reactants are of equal concentrations, but the volume of the base is half of that of the acid so the reaction has reached half of the volume of base needed for neutralisation. This means that the $\text{K}_a$ the reaction mixture is equal to the $\text{pH}$ of the reaction.

$\text{pH} =\text{pK}_a = -\text{log }\textcolor{#0040ff}{1.7\times10^{-5}} =\textcolor{#008d65}{4.77}$

A LevelAQA

## Example 5: Diluted Solutions

Calculate the new $\text{pH}$ of a solution when $\textcolor{#00bfa8}{55.0\text{ cm}^3}$ of $\textcolor{#f21cc2}{0.200\text{ mol dm}^{-3}}$ of $\text{NaOH}$ is mixed with $\textcolor{#a233ff}{750\text{ cm}^3}$ of water at$25\degree \text{C}$.

[7 marks]

\begin{aligned}\left[\text{OH}^-\right]&=\left[\text{OH}^-\right]_{\text{original}}\times\frac{\text{Initial Volume}}{\text{Final Volume}}\\ \text{}\\&=\textcolor{#f21cc2}{0.200}\times\frac{\textcolor{#00bfa8}{0.055}}{\textcolor{#a233ff}{0.750}}\\ \text{}\\ &=\textcolor{#008d65}{0.015\text{ mol dm}^{-3}}\end{aligned}

\begin{aligned}\text{K}_w&=\left[\text{H}^+\right]\left[\text{OH}^-\right]\\\text{}\\\left[\text{H}^+\right]&=\frac{\text{K}_w}{\left[\text{OH}^-\right]}\\ \text{}\\ &=\frac{10^{-14}}{0.015}\\ \text{}\\ &=\textcolor{#008d65}{6.67\times10^{-13}\text{ mol dm}^{-3}}\end{aligned}

$\text{pH}=-\text{log }7.32\times10^{-13}=\textcolor{#008d65}{12.2}$

A LevelAQA

## Neutralisations and Dilutions Example Questions

Step 1. Calculate the moles of $\text{H}^+$ ions.

$\text{Moles NaOH} = \text{Moles OH}^-=\frac{40}{1000}\times0.25=0.01\text{ mol}$

$\text{Moles of H}_2\text{SO}_4=\frac{30}{1000}\times0.18=5.4\times10^{-3}\text{ mol}$

$\text{Moles H}^+=5.4\times10^{-3}\times2=0.0108$

$\text{Excess Moles H}^+=8\times10^{1-^-4}$

Step 2: Calculate $\left[\text{H}^+\right]$.

$\left[\text{H}^+\right]=\frac{8\times10^{-4}}{0.07}=0.011\text{ mol dm}^{-3}$

Step 3: Calculate the $\text{pH}$.

$\text{pH}=1.9$

(Four marks for step 1, one mark per correct mole calculation. One mark for step 2. One mark for step 3.)

Step 1: Calculate the concentration of $\text{H}^+$.

$\left[\text{H}^+\right]=10^{-1.5}=0.032\text{ mol dm}^{-3}$

Step 2: Calculate the moles of $\text{HCl}$.

$\text{Moles HCl}=\frac{25}{1000}\times0.090=2.25\times10^{-3}$

Step 3: Calculate the final volume.

$\text{Final Volume}=\frac{2.25\times10^{-3}}{0.032}=0.070\text{ dm}^{-3} \left(71.2\text{ cm}^3\right)$

(one mark per correct step)

$\text{pH}=\text{pK}_a=-\text{log }1.51\times10^{-5}=4.8$

Step 1: Calculate the excess moles.

$\text{Moles Propanoic Acid}=\frac{75}{1000}\times0.80=0.06\text{ mol}$

$\text{Moles of NaOH}=\text{Moles of OH}^-=\frac{30}{1000}\times0.40=0.012\text{ mol}$

$\text{Excess Propanoic Acid} = 0.06-0.012=0.048\text{ mol}$

Step 2: Calculate the concentrations of the acid and the anion.

$\left[\text{Ethanoic Acid}\right]=\frac{0.048}{0.105}=0.457\text{ mol dm}^{-3}$

$\left[\text{Ethanoate Ions}\right]=\frac{0.012}{0.105}=0.114\text{ mol dm}^{-3}$

Step 3: Calculate the concentration of $\text{H}^+$.

\begin{aligned}\left[\text{H}^+\right]&=\frac{\text{K}_a\times\left[\text{CH}_3\text{CH}_2\text{COOH}\right]}{\left[\text{CH}_3\text{CH}_2\text{COO}^-\right]}\\ &=5.40\times10^{-5}\end{aligned}

Step 4: Calculate the $\text{pH}$.

$\text{pH}=-\text{log }5.40\times10^{-5}=4.3$

(Three marks for step 1, one mark per correct mole calculation. Two marks for step 2, one mark per correct concentration calculation. One mark for step 3. One mark for step 4.)

Step 1: Calculate the moles of $\text{H}^+$ in excess.

$\text{Moles H}_2\text{SO}_4=\frac{40}{1000}\times0.5=0.20\text{ mol}$

$\text{Moles H}^+=0.02\times2=0.04\text{ mol}$

$\text{Moles NaOH}=\text{Moles OH}^-=\frac{30}{1000}\times0.65=0.0195$

$\text{Excess H}^+=0.04-0.0195=0.0205\text{ mol}$

Step 2: Calculate $\left[\text{H}^+\right]$.

$\left[\text{H}^+\right]=\frac{0.0205}{0.07}=0.2929\text{ mol dm}^{-3}$

Step 3: Calculate the $\text{pH}$.

$\text{pH}=-\text{log }0.2929=0.53$

(Four marks for step 1, one mark per correct mole calculation. One mark for step 2. One mark for step 3.)

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