$\begin{matrix} List-I & List-II (Titrant required) (I) & K_{2} C_{2} O_{4} . N a H C_{2} O_{4} & (P) & N a O H 200 m L 1 M (0.2 mole) (I I) & 2 K_{2} C_{2} O_{4} . H_{2} C_{2} O_{4} & (Q) & C a (O H)_{2} 50 m L 2 M (0.1 mole) (I I I) & H_{2} C_{2} O_{4} . 2 N a H C_{2} O_{4} & (R) & K M n O_{4} (Acidic medium) (0.05 mole) & 0.2 M, 800 m L (S) & K M n O_{4} (Acidic medium) 0.2 M, 300 m L (T) & C a (O H)_{2} 500 m L 2 M (U) & N a O H 1000 m L 1 M \end{matrix}$

Which of the following options has the correct combination considering List-I and List-II.

I - P Q S

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I I - P Q

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I I I - R T U

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I - T U

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Solution

The correct option is B $I I - P Q$
$\begin{matrix} List-I & List-II (Titrant required) (A) & K_{2} C_{2} O_{4} . N a H C_{2} O_{4} & (P) & N a O H 200 m L 1 M (n_{f} = 1) 0.2 mole & meq base = 200 meq as acid = 200 (n_{f} = 1) meq as reducing agent = 800 & (Q) & C a (O H)_{2} 50 m L 2 M (n_{f} = 2) (n_{f} = 4) m_{e q} b a s e = 200 (B) & 2 K_{2} C_{2} O_{4} . H_{2} C_{2} O_{4} & (R) & K M n O_{4} (acid) (n_{f} = 5) 0.1 m o l e & 0.2 M, 800 m L meq as acid = 200 (n_{f} = 2) & m_{e q} o x i d i s i n g a g e n t = 800 meq as reducing agent 600 (n_{f} = 6) (C) & H_{2} C_{2} O_{4} . 2 N a H C_{2} O_{4} & (S) & K M n O_{4} (A c i d) (n_{f} = 5) (0.05 mole) & 0.2 M, V = 300 m L meq as acid = 200 (n_{f} = 4) & m_{e q} oxidising agent = 300 meq as reducing agent = 300 (n_{f} = 6) \end{matrix}$

$⎡ ⎢ ⎣ \begin{matrix} Acid + Base m_{e q} acid = m_{e q} base m_{e q} reducing agent = m_{e q} oxidising agent \end{matrix} ⎤ ⎥ ⎦$

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