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Salt of Strong Acid and Weak Base

The normality...

Question

The normality of a solution that results from mixing $4$ g of NaOH, $500$ ml of $1$ M HCl, and $10.0$ ml of $H_{2} O_{4}$ (specific gravity $1.1$ , $49$ % $H_{2} S O_{4}$ by weight) is :
(The total volume of solution was made to $1$ L with water)

0.51

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0.71

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1.02

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0.45

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Solution

The correct option is A 0.51
$4$ g of $N a O H = \frac{4}{40} = 0.1 m o l e = 100 m m o l \equiv 100 m E q$
$H C l = 500 \times 1 = 500 m E q$
Normality of $H_{2} S O_{4} = \frac{W_{2} \times 1000}{E w_{2} \times V_{s o l} (i n m L)}$
$= \frac{percentage by weight \times 10 \times d}{E w_{2}}$
$= \frac{49 \times 10 \times 1.1}{49} = 11 N$
mEq of $H_{2} S O_{4} = 11 N \times 10.00 m L = 110 m E q$
Total acid $= 110 + 500 = 610 m E q$
NaOH $= 100 m E q$
Acid left $= 610 - 100 = 510 m E q$
Total volume $= 1000 m L$
Normality of solution $= \frac{m E q}{m L} = \frac{510}{1000} = 0.51 N$

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