The **question aims** to find the **pH** of a buffer.

The **measure of acidity or authenticity of aqueous or other liquid solutions** is defined as **pH**. This **term** is normally used in chemistry, biology, and agronomy, and translates concentrations of hydrogen ions — usually between **1** and **10−14** per gram per liter — into numbers between **0** and **14**.

A simple buffer solution contains an** acid solution** and salt **conjugate base acid**. **For example**, **acid can be acetic**, and **salt** can be s**odium acetate**. The **Henderson Hasselbalch** calculator associates the $pH$ of a solution consisting of a mixture of two particles with the stability of acid separation, the $Ka$ of acid, and the **concentration** of the solution type.

**Following simplifying assumptions are used to derive the equation.**

**Assumption 1:** Acid, $HA$, **monobasic** and **differentiates** according to the equation.

\[HA\rightleftharpoons H^{+}+A^{-}\]

\[C_{a}=[A^{-}]+\dfrac{[H^{+}][A^{-}]}{K_{a}}\]

\[C_{H}=[H^{+}]+\dfrac{[H^{+}][A^{-}]}{K_{a}}\]

$C_{a}$ is the **concentration of acid** analysis and $CH$ is the **concentration of hydrogen ion** that has been added to the solution.

The **Henderson Hasselbalch** scale can only be used in polybasic acid if its consecutive $pH$ values vary by at least $3$. **Phosphoric acid is such an acid.**

**Assumption 2:** **Water self-ionizatio**n can be overlooked. This argument is not, at present, permissible with $pH$ values close to $7$, half the $pK_{w}$ value, which is a constant **ionization of water**. In this case, the **mass-balance equation** of hydrogen should be extended to consider** water ionization.**

\[C_{H}=\dfrac{[H^{+}][A^{-}]}{K_{a}}+\dfrac{K_{w}}{H^{+}}\]

**Assumption 3:** **Salt** $MA$ is completely **separated from the solution.** **For example, sodium acetate**

\[Na(CH_{3}CO_{2}\rightarrow Na^{+}+CH_{3}CO_{2}^{-} \]

**saturation of sodium ion**, $[Na ^{+}]$ is ignored. This is a good ratio for $1: 1$ electrolyte, but not ion salts with a high charge like **magnesium sulphate**, $Mg (SO_{4})_{2}, which makes ion pairs.

**Assumption 4:**

Value of the $K_{a}$

\[K_{a}=\dfrac{[H^{+}][A^{-}]}{HA}\]

**Rearrangement** of this **equation and logarithm** provision gives the **Henderson Hasselbalch equation:**

\[pH=pK_{a}+\log\dfrac{A^{-}}{HA}\]

The **Henderson-Hasselbalch equation** is used to find the $pH$ of the solution.

## Expert Answer

Using** Henderson-Hasselbalch equation:**

\[pH=pK_{a}+\log\dfrac{A^{-}}{HA}\]

$HA(CH_{2}CHOHCOOH)$ is the acid $A^{-}(CH_{2}CHOHCOONA)$ is its conjugate base.

$pK_{a}$ is given, which is **acid strength.**

\[pK_{a}=3.86\]

The **acid value** is given as:

\[CHOHCOOH=0.12 M\]

The** conjugate base** is given as:

\[CHOHCOONA=0.11 M\]

**Plug** the values into the **Henderson-Hasselbalch equation** to calculate the $pH$.

\[pH=3.86+\log\dfrac{0.11}{0.12}\]

\[pH=3.822\]

Hence, $pH$ is $3.822$.

## Numerical Result

**Buffer** that has $pH$ $0.12$ $M$ in **lactic acid** and $0.11$ $M$ in **sodium lactate** is **calculated** as:

\[pH=3.822\]

## Example

**Find the $pH$ of a buffer that is $0.15$ $M$ in lactic acid and $0.17$ $M$ in sodium lactate.**

**Henderson-Hasselbalch equation** is used to find the $pH$ of the** solution. **

\[pH=pK_{a}+\log\dfrac{A^{-}}{HA}\]

$HA(CH_{2}CHOHCOOH)$ is the **acid** $A^{-}(CH_{2}CHOHCOONA)$ is its **conjugate base.**

$pK_{a}$ is shown below, which is **acid strength.**

\[pK_{a}=3.86\]

The **acid value** is given as:

\[CHOHCOOH=0.15 M\]

The **conjugate base** is given as:

\[CHOHCOONA=0.17 M\]

**Plug** the values into the **Henderson-Hasselbalch** equation to find the $pH$.

\[pH=3.86+\log\dfrac{0.17}{0.15}\]

\[pH=3.914\]

**Buffer** with $0.15$ $M$ in** lactic acid** and $0.17$ $M$ in **sodium lactate** has $pH$ **calculated** as:

\[pH=3.914\]

Hence, $pH$ is $3.914$.