NaHC204 is 0.1 M when neutralized with NaOH. Hence, it is .. when oxidized with MnO /H.
(A)0.1 N
(B) 0.2 N
(C) 0.05 N
(D)0.15 N
Answers
Answer:
NaHC
2
O
4
⇌Na
+
+
0.1↑neutralisedby OH
−
H
+
+
0.1↑oxidized byMnO
4
−
C
2
O
4
2−
0.1 mol of sodium oxalate on dissociation gives 0.1 mole of hydrogen ions and 0.1 mole of oxalate ions. 0.1 mole of hydrogen ions is neutralized by 0.1 mole of sodium hydroxide. 1 mole of calcium hydroxide neutralizes 2 moles of hydrogen ions.
Hence, 0.1 mole of hydrogen ions are neutralized by 0.5 moles of calcium hydroxide.
0.1 mol
1H
+
≡
0.1 mol
1OH
−
≡
0.1 mol
NaOH
≡
0.05 mol
2
1
Ca(OH)
2
Hence, options (A) and (B) are true.
In acidic medium, the reaction between oxalate ions and permanganate ions is 5C
2
O
4
2−
+2MnO
4
−
⟶2Mn
2+
+10CO
2
.
Hence, 5 moles of oxalate ions are oxidized by 2 moles of permanganate ions. Hence, 0.1 moles of oxalate ions will be oxidized by
5
2
×0.1=0.04 moles of potassium permanganate.
Thus, the option (C) is also correct.
In the basic medium, C
2
O
4
2−
≡2MnO
4
−
.
Hence, 1 mole of oxalate ions is oxidized by 2 moles of permanganate ions and 0.1 moles of oxalate ions will be oxidized by
1
2
×0.1=0.2 moles of potassium permanganate.
Thus, the option (D) is incorrect.
Answer:
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