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elena55 [62]
2 years ago
6

A titration process is done by using 0.2M of 20 ml of third analytical group main reagent solution with 0.1 M nitric acid soluti

on if nitric acid is replaced by sulphuric acid of same concentration, then the volume of sulphuric acid equals ……
Half the volume of nitric acid
Double the volume of nitric acid
The volume of nitric acid
Quarter the volume of nitric acid
Chemistry
1 answer:
Pavel [41]2 years ago
8 0

Sulfuric acid undergoes dissociation reaction to produce twice the amount

of hydrogen ions produced by nitric acid in solution.

  • The volume of sulfuric acid equals; <u>Half the volume of nitric acid</u>

Reasons:

The concentration of the analytical group = 0.2 M

Volume of reagent = 20 ml

Concentration of the nitric acid to be replaced = 0.1 M

Replacement for the nitric acid = Sulfuric acid

Concentration of the sulfuric acid = Concentration of the nitric acid = 0.1 M

Reaction = Titration

Solution:

Sulfuric acid in solution gives; H₂SO₄ (aq)  ⇄ 2H⁺ (aq) + SO₄²⁻(aq)

For nitric acid, we have; HNO₃ (aq) ⇄ H⁺(aq) + NO₃⁻(aq)

Therefore;

1 mole of sulfuric acid gives 2 moles of hydrogen ions

0.1 M solution of sulfuric acid gives 0.2 moles of hydrogen ions per liter

0.1 moles of hydrogen ions are therefore present in half a liter solution of 0.1 M sulfuric acid solution.

However;

0.1 M of nitric acid gives 0.1 moles of hydrogen ions per liter

Therefore;

  • The volume of sulfuric acid required = <u>Half the volume of nitric acid</u>

Learn more about titration and quantitative chemical analysis here:

brainly.com/question/6800262

brainly.com/question/6548806

brainly.com/question/13031875

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Read 2 more answers
Consider the reaction 2 NO + O2 → 2 NO2
Sonbull [250]

Answer:

(a) Rate at which NO_2 is formed is 0.050 M/s

(b) Rate at which O_2 is consumed is 0.0250 M/s.

Explanation:

The given reaction is:-

2NO+O_2\rightarrow 2NO_2

The expression for rate can be written as:-

-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}

Given that:- \frac{d[NO]}{dt}=-0.050\ M/s (Negative sign shows consumption)

-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}

-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}

-(-0.050\ M/s)=\frac{d[NO_2]}{dt}

\frac{d[NO_2]}{dt}=0.050\ M/s

(a) Rate at which NO_2 is formed is 0.050 M/s

-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}

-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}

\frac{d[O_2]}{dt}=0.0250\ M/s

(b) Rate at which O_2 is consumed is 0.0250 M/s.

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