$2 C H_{3} C O O K + 2 H_{2} O ⟶ C_{2} H_{6} + 2 C O_{2} + 2 K O H + H_{2}$

During the above reaction, the pH of the solution:

increases

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decreases

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remains constant

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cannot be predicted

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Solution

The correct option is A increases
$2 C H_{3} C O O K + 2 H_{2} O ⟶ C_{2} H_{6} + 2 C O_{2} + 2 K O H + H_{2}$

The given process is Kolbe's electrolysis for preparation of alkane. In this reaction, sodium or potassium salt of carboxylic acid undergoes electrolysis to form alkane and sodium or potassium hydroxide.

Because of the formation of sodium or potassium hydroxide (strong bases), the pH of the solution increases.

Hence, option $A$ is correct.

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Similar questions

2 C H_{3} C O O K + 2 H_{2} O ⟶ C_{2} H_{6} + 2 C O_{2} + 2 K O H + H_{2}

C H_{3} C O O^{-} N a^{+} + 2 H_{2} O \to C H_{3} C H_{3} + 2 C O_{2} + H_{2} + 2 N a O H

C l_{2} + 2 N a B r \to B r_{2} + 2 N a C l

2 N a + O_{2} \to N a_{2} O

2 K + 2 H_{2} O \to H_{2} + 2 K O H

2 H_{2} O ⇌ 2 H_{2} + O_{2} K_{1} = 2.1 \times 10^{13}

2 C O_{2} ⇌ 2 C O + O_{2} K_{2} = 1.4 \times 10^{12}

H_{2} + C O_{2} ⇌ C O + H_{2} O

2 K + 2 H_{2} O \to 2 K O H + H_{2}

2.0 m o l e s

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