Question

A solution of sodium hydroxide 0.2 mol/dm³ was neutralised by 20 cm³ of sulphuric acid 0.1 mol/dm³ in reaction of neutralisation. Find the volume of sodium hydroxide that was used in the reaction.​

Answer 1

Answer:

Explanation:

Write the BALANCED equation for the reaction:

2 NaOH  + H2SO4  --->  Na2SO4  +  2 H2O

You’ll see that 1 mole H2SO4 requires 2 moles NaOH for neutralisation

50cm^3 of Sulphuric acid, concentration 0.1 mol/l

Contains:

0.1 x 50 divided by 1000 moles H2SO4

This is equivalent to:

0.1 x 50 x 2 divided by 1000 moles NaOH

= 0.01 moles NaOH

Sodium Hydroxide solution, concentration 0.4 mol/l contains  

0.4 divided by 1000 moles NaOH per cm^3

Therefore 0.01 moles NaOH is equivalent to:

1000 x 0.01 divided by 0.4 cm^3 NaOH solution

= 25 cm^3

OR, another way:

Write the BALANCED equation for the reaction:

2 NaOH  + H2SO4  --->  Na2SO4  +  2 H2O

Since 1 mole H2SO4 requires 2 mole NaOH,

50cm^3 of Sulphuric acid, concentration 0.1 mol/l will require

2 x 50cm^3 NaOH, concentration 0.1 mol/l

Therefore, amount of 0.4 mol/l NaOH required is:

2 x 50 divided by 4

= 25 cm^3

Answer 2

Answer:

20cm³

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

From the equation above,

nA (mole of acid) = 1

nB (mole of the base) = 2

Data obtained from the question include:

Ma = 0.1 mol/dm³

Va = 20cm³

Mb = 0.2 mol/dm³

Vb =?

Using MaVa /MbVb = nA/nB, the volume of the base can easily be obtained as follows:

MaVa = MbVb

0.1 x 20 / 0.2 x Vb = 1/2

Cross multiply to express in linear form.

1 x 0.2 x Vb = 0.1 x 20 x 2

Divide both side by 0.2

Vb = (0.1 x 20 x 2) / 0.2

Vb = 20cm³

Therefore, the volume of the NaOH required is 20cm³