Question

In Walker Formal method following observation had been obtained for a given sample of milk
1. mL of milk taken: 10
2. Burette reading after 1st titration: 2.5
3. Burette reading after 2nd titration: 3.4
What is the % total protein and % casein content of the milk sample ?

• A. % total protein:1.53 & % casein: 1.206
• B. % total protein: 1.75 & %casein: 1.5
• C. % total protein: 2.5 & % casein: 2
• D. % total protein: 3.125 & % casein: 2.5

Hint:

In Formol titration of \(10\text{ mL}\) milk, the difference between the two titration readings (Formol Titre) is multiplied by \(1.70\) to obtain % Total Protein, and by \(1.34\) to obtain % Casein.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The Walker Formol titration is a rapid volumetric method used to estimate the casein and total protein content in milk.
The addition of formaldehyde to neutralized milk releases hydrogen ions from the amino groups of proteins, which can then be titrated with a standard alkali solution.
Key Formula or Approach:
The Formol Titre represents the volume of standard sodium hydroxide (\(0.1\text{ N}\)) used in the second titration:
\[ \text{Formol Titre} = \text{Burette Reading 2} - \text{Burette Reading 1} \]
For a \(10\text{ mL}\) milk sample:
- % Total Protein = \(\text{Formol Titre} \times 1.70\)
- % Casein = \(\text{Formol Titre} \times 1.34\)

Step 2: Detailed Explanation:
Let us perform the calculations using the given data:
-
Step 1: Calculate the Formol Titre:
\[ \text{Formol Titre} = 3.4\text{ mL} - 2.5\text{ mL} = 0.9\text{ mL} \]
- Calculate the % Total Protein using the standard conversion factor of \(1.70\) for a \(10\text{ mL}\) sample:
\[ \% \text{Total Protein} = 0.9 \times 1.70 = 1.53\% \]
-
Step 2: Calculate the % Casein using the standard Walker conversion factor of \(1.34\) for a \(10\text{ mL}\) sample:
\[ \% \text{Casein} = 0.9 \times 1.34 = 1.206\% \]
Alternatively, since casein makes up approximately \(78.8\%\) of the total milk protein:
\[ \% \text{Casein} = 1.53\% \times 0.788 \approx 1.206\% \]
These values match option (A).

Step 3: Final Answer:
The milk sample contains 1.53% total protein and 1.206% casein.