Question

Which of the statements is not true?

• A. On passing $H_2S$ through acidified $K_2Cr_2O_7$ solution, a milky colour is observed
• B. $Na _2Cr_2O_7$ is preferred over $K_2Cr_2O_7$ in volumetric analysis
• C. $K_2Cr_2O_7$solution in acidic medium is orange
• D. $K_2Cr_2O_7 $ solution becomes yellow on increasing the pH beyond 7

Answer 1

The Correct Option is B

To determine which statement is not true, we need to analyze each option carefully with respect to chemical properties and practices:

1. On passing \(H_2S\) through acidified \(K_2Cr_2O_7\) solution, a milky colour is observed: 
   • This statement is true because hydrogen sulfide (\(H_2S\)) reacts with acidified potassium dichromate (\(K_2Cr_2O_7\)) to form elemental sulfur, which is a solid and gives a milky appearance. The chemical equation is: \(K_2Cr_2O_7 + 3H_2S + 4H_2SO_4 \rightarrow 3S + Cr_2(SO_4)_3 + K_2SO_4 + 7H_2O\).
2. \(Na _2Cr_2O_7\) is preferred over \(K_2Cr_2O_7\) in volumetric analysis:
   • This statement is not true. In practice, \(K_2Cr_2O_7\) (potassium dichromate) is more commonly used in volumetric analysis due to its higher purity and stability compared to \(Na _2Cr_2O_7\) (sodium dichromate). Potassium salts are generally preferred due to their better handling properties.
3. \(K_2Cr_2O_7\) solution in acidic medium is orange:
   • This statement is true. \(K_2Cr_2O_7\) appears orange in acidic conditions because of the presence of dichromate ions, which absorb light to produce an orange solution.
4. \(K_2Cr_2O_7\) solution becomes yellow on increasing the pH beyond 7:
   • This statement is true. As the pH increases (in alkaline conditions), \(K_2Cr_2O_7\) converts to chromate ions, which are yellow in color. The equilibrium shift from dichromate ions to chromate ions causes the color change: \(2CrO_4^{2-} + 2H^+ \leftrightarrows Cr_2O_7^{2-} + H_2O\).

Thus, the statement that is not true is: \(Na _2Cr_2O_7\) is preferred over \(K_2Cr_2O_7\) in volumetric analysis.