Question

2. 10 ml of N/30 acid is titrated against a base using phenolphthalein as indicator. If 7 ml of base is used for titration find out the Normality of the base.​

Answer 1

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When phenolphthalein is the indicator, whole of NaOH has been neutralised and carbonate converted into bicarbonate, i.e.,

NaOH+HCl→NaCl+H2 ONa 2 CO 3 +HCl→NaHCO3 +NaCl

So, 25 mL 10 HCl≡NaOH+1/2Na 2

CO

3

present in 25 mL of mixture.

In another titration, when methyl orange is the indicator, whole of NaOH has been neutralised and carbonate converted into carbonic acid, i.e.,

Na

2

CO

2

+2HCl→2NaCl+H

2

CO

3

30mL

10

N

HCl≡NaOH+Na

2

CO

3

present in 25 mL of mixture

Hence,

(30−25)mL

10

N

HCl≡

2

1

Na

2

CO

3

present in 25 mL of mixture

Hence,

10 mL

10

N

HCl≡Na

2

CO

3

present in 25 mL of mixture

≡10 mL

10

N

Na

2

CO

3

solution

Amount of Na

2

CO

3

=

10×1000

53×10

=0.053 g

This amount of Na

2

CO

3

is present in 25 mL of mixture.

This amount present in one litre of mixture,

=

25

0.053

×1000=2.12 g

(30−10)mL

10

N

HCl≡NaOH present in 25 mL of mixture

≡20 mL

10

N

NaOH

Amount of NaOH in 25 mL of mixture =

10×1000

40×20

=0.08 g

The amount present in one litre of mixture =

25

0.08

×1000=3.20 g.

Answer 2

Answer:

The normality of the base is N/21.

Explanation:

Neutralization of a solution is possible when the amount of base in equivalent proportion to the amount of acid.

Therefore, at neutralization, the equivalent mass of acid dissolved in volume $V_1$ of solvent should be equal to the equivalent mass of the base dissolved in volume $V_2$ of solvent.

The number of gram equivalent mass of solute dissolved in one litre of the solution is called the normality of the solution

If $N_1$ and $N_2$ are equivalent masses of acid and base dissolved in the solution of 1 liter, then $N_1V_1 = N_2V_2$

Given the volume of acid, $V_1=10 ml$

Equivalent mass of acid,

$N_1=\dfrac{N}{30}$  

The volume of base, $V_2=7ml$

Then equivalent mass of base is  

$N_2=\dfrac{N_1\times V_1}{V_2}$

    $=\dfrac{\dfrac{N}{30} \times 10}{7}=\dfrac{N}{3} \times \dfrac{1}{7}=\dfrac{N}{21}$  

Therefore, the normality of the base is N/21.

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