Q19.129CP - Chemistry: Molecular Nature Of Matter And Change

Question

Muscle physiologists study the accumulation of lactic acid $\left[{\text{CH}}_3\text{CH}\left(\text{OH}\right)\text{COOH}\right]$ during exercise. Food chemists study its occurrence in sour milk, beer, wine, and fruit. Industrial microbiologists study its formation by various bacterial species from carbohydrates. A biochemist prepares a lactic acid-lactate buffer by mixing  $\mathbf{225}\mathbf{ }\mathbf{ml}$ of $\mathbf{0}\mathbf{.}\mathbf{85}\mathbf{ }\mathbf{M}$  lactic acid  $\left(K_a=1.38\times {10}^{-4}\right)$ with $\mathbf{435}\mathbf{ }\mathbf{mL}$ of  $\mathbf{0}\mathbf{.}\mathbf{68}\mathbf{ }\mathbf{M}$ sodium lactate. What is the buffer  ?

Step-by-Step Solution

Answer

The required buffer pH is 4.05.

Buffer solution

Buffer solutions contain weak acids in equilibrium with their conjugate bases. Buffer solutions do not show changes in pH after addition of acid or base. The Henderson – Hasselbalch equation is very helpful in estimation of pH of a buffer solution. For a weak acid HA in equilibrium with its conjugate base A-, the Henderson-Hasselbalch equation is given as :

$\mathbf{pH}\mathbf{=}\mathbf{pKa}\mathbf{+}\mathbf{log}\frac{\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}\mathbf{HA}\mathbf{\right]}}$

Calculate pH

Volume of lactic acid $=\text{225} \text{ml} \text{ = } \text{0.225} \text{L}$  

Concentration of Lactic acid$=\text{0.85} \text{M}$   

Volume of sodium lactate $=435 \mathrm{ml} = 0.435 \mathrm{L}$

Concentration of Lactic acid $=\text{0.68} \text{M}$ 

 ${\mathrm{K}}_{\mathrm{a}}=1.38\times {10}^{-4}$

To calculate pH of a buffer solution, we just have to use Henderson-Hasselbalch equation:

$\mathrm{pH}=\mathrm{pKa}+\log \frac{\left[\mathrm{A}\right]}{\left[\mathrm{HA}\right]}$ .

Instead of using concentrations, in the Henderson-Hasselbalch equation, we can also use the number of moles of Lactic acid and sodium lactate present in the buffer solution.

$$\begin{gathered} \mathrm{pH}=-\log \left(1.38\times {10}^4\right)+\log \frac{\left(0.435\mathrm{L}\times 0.68\mathrm{M}\right)}{\left(0.225\mathrm{L}\times 0.85\mathrm{M}\right)} \\ \mathrm{pH}=4.05 \end{gathered}$$

Therefore, the required pH value is $4.05$.