You have 0-490 g of unknown acid- H2A- which reacts with KOH- according to following reaction- H2Aaq - 2KOH aq - K2A aq - 2H2O l-If 100 mL of 0-1M KOH is required to titrate the acid to the second equivalence point- then the molar mass of acid is -

Let E be the equivalent mass of the acid.The number of equivalents nbsp;of nbsp;the acid nbsp;is nbsp; 0.490E .The number of equivalents o ...

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Standard XII

Chemistry

Stoichiometry

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Question

You have $0.490$ g of unknown acid, $H_{2} A$ , which reacts with $K O H$ , according to following reaction:
$H_{2} A (a q) + 2 K O H (a q) \to K_{2} A (a q) + 2 H_{2} O (l)$ .
If $100$ mL of $0.1 M$ $K O H$ is required to titrate the acid to the second equivalence point, then the molar mass of acid is :

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Solution

Let $E$ be the equivalent mass of the acid.
The number of equivalents of the acid is $\frac{0.490}{E}$ .
The number of equivalents of $K O H$ is $\frac{100 \times 0.1}{1000}$
The number of equivalents of the acid is equal to the number of equivalents of $K O H$ .
$⟹ \frac{0.49}{E} = 0.01$
Thus, the equivalent mass of the acid is $49 g m o l^{- 1}$ .
For the dibasic acid, the molar mass is twice its equivalent mass. It is equal to $98 g m o l^{- 1}$ .

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You have 0.490 g of unknown acid, H2A, which reacts with KOH, according to following reaction: H2A(aq)+2KOH(aq)→K2A(aq)+2H2O(l).If 100 mL of 0.1M KOH is required to titrate the acid to the second equivalence point, then the molar mass of acid is :