Analytical Chemistry Question 167
Question: $ KMnO_4 $ reacts with oxalic acid as:
$ MnO_4^{-}+C_2O_4^{2-}+{H^{+}}\xrightarrow{{}}M{n^{2+}}+CO_2+H_2O $ Hence, 50 ml of 0.04 M $ KMnO_4 $ is acidic medium is chemically equivalent to
Options:
A) 100 ml of 0.1M $ H_2C_2O_4 $
B) 50 ml of 0.2 M $ H_2C_2O_4 $
C) 50 ml of 0.1M $ H_2C_2O_4 $
D) 25 ml of 0.1M $ H_2C_2O_4 $
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Answer:
Correct Answer: C
Solution:
[c] Equiv. mass of $ KMnO_4 $
$ MnO_4^{-}=\frac{molarmass}{7-2}=\frac{molarmass}{5} $ Equiv. mass of oxalic acid $ C_2O_4^{2-}=\frac{molarmass}{2(4-3)}=\frac{molarmass}{2} $ Meq. Of $ KMnO_4=50\times 5\times 0.04=10= $ meq of $ H_2C_2O_4=50\times 2\times 0.1=10. $ Hence [c].
BETA
