Some Basic Concepts of Chemistry - Result Question 96
####34. An aqueous solution containing $0.10 g KIO _3$
$($ formula weight $=214.0)$ was treated with an excess of KI solution. The solution was acidified with $HCl$. The liberated $I _2$ consumed $45.0 mL$ of thiosulphate solution decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution.
$(1998,5 M)$
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Solution:
- The balanced equations are
(1) $2 MnCl _2+5 K _2 S _2 O _8+8 H _2 O \rightarrow$
$$ 2 KMnO _4+4 K _2 SO _4+6 H _2 SO _4+4 HCl $$
(2) $2 KMnO _4+5 H _2 C _2 O _4+3 H _2 SO _4 \rightarrow$
$$ K _2 SO _4+2 MnSO _4+8 H _2 O+10 CO _2 $$
Given, mass of oxalic acid added $=225 mg$
So, millimoles of oxalic acid added $=\frac{225}{90}=2.5$
Now from equation 2
Millimoles of $KMnO _4$ used to react with oxalic acid=1 and Millimoles of $MnCl _2$ required initially $=1$
$\therefore$ Mass of $MnCl _2$ required initially $=1 \times(55+71)=126 mg$
Alternative Method
$m$ moles of $MnCl _2=m$ moles of $KMnO _4=x$ (let)
and $M _{eq}$ of $KMnO _4=M _{eq}$ of oxalic acid
So, $\quad x \times 5=\frac{225}{90} \times 2$
Hence, $x=1$
$\therefore \quad m$ moles of $MnCl _2=1$
Hence mass of $MnCl _2=(55+71) \times 1=126 mg$.
