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.

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

the mass of the cathode increases because ions from the solution get reduced to atoms and become part of the cathode.

explanation:

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