In iodometric titrations- an oxidizing agent such as KMnO 4- K 2 Cr 2 O 7- CuSO 4- H 2 O 2 is allowed to react in neutral medium or in acidic medium with excess of potassium iodide to liberate free iodine I 2- Free iodine is titrated against standard reducing agent usually with sodium thiosulphate- The reactions are as follows-K 2 Cr 2 O 7-6 KI -7 H 2 SO 4- Cr 2 SO 43-4 K 2 SO 4-7 H 2 O - I 22 CuCO 4-4 KI - Cu 2 I 2-2 K 2 SO 4- I 2I 2- N a 2 S 2 O 3- 2 N aI - N a 2 S 4 O 6In iodometric titrations- starch solution is used as an indicator- Starch solution gives blue or violet colour with free iodine- At the end point- blue or voilet colour disappears when iodine is completely changed to iodide-What volume of 0-40 M N a 2 S 2 O 3 would be required in mL to react with I 2 liberated by adding 0-04 mole of KI to 50 mL of 0-20 M CuSO 4 solution-

Meq. of KI≡ meq. of CuSO_4≡ meq. of I_2 meq≡ meq. of Na_2S_2O_3 mequivalent of KI = 0.04× 10^3 = 40 mequivalent of CuSO_4 = 50× 0.2 = 10 nbsp; ...

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Equivalent Mass

In iodometric...

Question

In iodometric titrations, an oxidizing agent such as $K M n O_{4}, K_{2} C r_{2} O_{7}, C u S O_{4}, H_{2} O_{2}$ is allowed to react in neutral medium or in acidic medium with excess of potassium iodide to liberate free iodine $(I_{2})$ .
Free iodine is titrated against standard reducing agent usually with sodium thiosulphate. The reactions are as follows:
$K_{2} C r_{2} O_{7} + 6 K I + 7 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 4 K_{2} S O_{4} + 7 H_{2} O + I_{2}$
$2 C u C O_{4} + 4 K I \to C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}$
$I_{2} + N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}$
In iodometric titrations, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine. At the end point, blue or voilet colour disappears when iodine is completely changed to iodide.

What volume of 0.40 M $N a_{2} S_{2} O_{3}$ would be required (in mL) to react with $I_{2}$ liberated by adding 0.04 mole of $K I$ to 50 mL of 0.20 M $C u S O_{4}$ solution?

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(a)

C u S O_{4}

K I

I_{2}

2 C u S O_{4} + 4 K I ⟶ C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}

(b)

H g_{5} (I O_{6})_{2}

K I

H C l

H g_{5} (I O_{6})_{2} + 34 K I + 24 H C l ⟶ 5 K_{2} H g I_{4} + 8 I_{2} + 24 K C l + 12 H_{2} O

(c)

N a_{2} S_{2} O_{3}

0.0499

C u S O_{4} .5 H_{2} O

N a_{2} S_{2} O_{3}

I_{2}

0.7245

H g_{5} (I O_{6})_{2}

C u S O_{4} .5 H_{2} O = 249.5

H g_{5} (I O_{6})_{2} = 1448.5

2 C u^{2 +} + 4 I^{-} \to C u_{2} I_{2} + I_{2}; I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}

0.12 m o l e

C u S O_{4}

K I

120 m L

0.5 g

K_{2} C r_{2} O_{7}

K M n O_{4}

K I

100 c m^{3}

0.15 N

0.5

K_{2} C r_{2} O_{7}

K M n O_{4}

K I

150 c m^{3}

0.10 N

K_{2} C r_{2} O_{7}

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