The aim is to explain the concepts of **pH** and **pOH** values. How to find and **calculate** the pH values by seeing the amount of **water** in the **solution.**

pH is a **measurement** of how **acidic/basic** the water is. The range varies from** 0 – 14,** where 7 denotes the **neutral.** The solution is said to be **acidic** if the **pH** is less than 7, whereas a **pH** of greater than 7 denotes a **base.**

pH is a measure of the comparative **quantity** of free **hydroxyl** and **hydrogen** ions in the water. **Acidic** water is the water that has more free **hydrogen** ions, whereas water with more free **hydroxyl** ions is denoted as basic. Since **chemicals** can affect the **pH** in water, pH is a significant indicator of water that is altered **chemically.**

pH is documented in **“logarithmic units”.** Each number describes a **10-fold** difference in the **acidity/basicity** of water. A pH of 5 is **ten** times less **acidic** than water having a pH of 4. The formula for calculating **pH** is given as:

\[ pH = -log[H^+] \]

pOH is a **measurement** of the **hydroxide** ion (OH^{–}) concentration of a solution. **expressions** of poH are:

\[pOH=14-pH\]

\[pOH = -log[OH^-] \]

## Expert Answer

First, convert the volume of **ethylamine** $20 mL$ into Liters $L$:

\[\dfrac{20}{1000} = 0.02 L\]

The **concentration** of ethylamine is given as $0.150$, The number of **moles** ($n$) of ethylamine can be **calculated** as:

\[ n_{eth} \space = 0.150 \space mol \times 0.02 \space L\]

\[ n_{eth} \space = 3.0 \times 10^{-3} \space mol \]

Converting the **volume** of HCI $50 \space mL$ to Liters $L$.

\[\dfrac{5}{1000} = 0.005 L\]

The **concentration** of HCI is given as $0.981 \space mol$ and the number of **moles** of HCI ($n$) can be **calculated** as:

\[n_{HCI} \space = 0.0981 \space mol \times 0.005 \space L\]

\[n_{HCI} \space = 4.91 \times 10^{-4} \space mol\]

After the **addition** of $5.0 mL$ of $0.981M$ of HCI in $20.0 mL$ of $0.150M$ of **ethylamine,** the pOH (Power of Hydroxide) is **calculated** as:

\[ pOH \space = \space pK_b + log \left( \dfrac{(n_{HCI}) / 25mL}{(n_{eth})/25mL} \right) \]

\[ = \space 3.25 + log \left( \dfrac{(4.91 \times 10^{-4}) / 25mL}{(3.0 \times 10^{-3})/25mL} \right) \]

\[ = 3.25 – 0.786\]

\[ pOH =2.46 \]

Now, the formula for **calculating** pH is given:

\[pH + pOH = 14 \]

**Rearranging** and making pH the subject:

\[ pH = 14 – pOH \]

**Inserting** the value of pOH and solving for pH:

\[ pH = 14 – 2.46 \]

\[ pH = 11.54 \]

## Numerical Answer

**pH** after the addition of $5.0 mL$ of HCl is $11.54$

## Example

Calculate the pH of **ethanoic acid** at $ 298K$ when the **hydrogen** concentration is $ 1.32 \times 10^{-3} mol dm^3$.

The formula for finding **pH** is given as:

\[ pH = -log[H^+] \]

\[ = -log(1.32 \times 10^{-3}) \]

\[ pH = 2.9 \]

**pH** of **ethanoic acid** at $ 298K$ is $2.9 $ when the **hydrogen concentration** is $ 1.32 \times 10^{-3} mol \space dm^3$.