A definite amount of BaCl2 was dissolved in HCl solution of unknown normality- 20 mL of this solution was treated with 21-4 mL of N10NaOH for complete neutralization- Further 20 mL of solution was added to 50 mL of N10Na2CO3 and the precipitate was filtered off- The filtrate reacted with 10-5 mL of 0-08 N H2SO4 using phenolphthalein as an indicator- The strength of HCl g-L in the mixture is as nearest integer -

Meq. of HCl added in BaCl_2= Meq. of NaOH used 20× N=21.4×110=2.14 N_HCl =0.107 Strength of HCl =0.107× 36.5 #160; = 3.90 ...

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Standard XII

Chemistry

Types of Redox Reactions

A definite am...

Question

A definite amount of $B a {C l}_{2}$ was dissolved in $H C l$ solution of unknown normality, $20$ mL of this solution was treated with $21.4$ mL of $\frac{N}{10} N a O H$ for complete neutralization. Further $20$ mL of solution was added to $50$ mL of $\frac{N}{10} {N a}_{2} C O_{3}$ and the precipitate was filtered off. The filtrate reacted with $10.5$ mL of $0.08 N H_{2} S O_{4}$ using phenolphthalein as an indicator. The strength of $H C l$ (g/L) in the mixture is (as nearest integer) :

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Solution

Meq. of $H C l$ added in $B a {C l}_{2} =$ Meq. of $N a O H$ used
$20 \times N = 21.4 \times \frac{1}{10} = 2.14$
$N_{H C l} = 0.107$
Strength of $H C l = 0.107 \times 36.5$ $= 3.9055$ g/L
The nearest integer is $4$ .

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Meq. of HCl added in BaCl2= Meq. of NaOH used20×N=21.4×110=2.14NHCl=0.107Strength of HCl=0.107×36.5 =3.9055 g/LThe nearest integer is 4.

Meq. of $H C l$ added in $B a {C l}_{2} =$ Meq. of $N a O H$ used
$20 \times N = 21.4 \times \frac{1}{10} = 2.14$
$N_{H C l} = 0.107$
Strength of $H C l = 0.107 \times 36.5$ $= 3.9055$ g/L
The nearest integer is $4$ .