The volume in cm3 3 of a 0.12 moldm−3 − 3 HCl required to completely neutralize a 20cm3 3 of 0.20 moldm−3 − 3 of NaOH is

The volume in cm3 ${}^3$ of a 0.12 moldm−3 ${}^{-3}$ HCl required to completely neutralize a 20cm3 ${}^3$ of 0.20 moldm−3 ${}^{-3}$ of NaOH is 

Answer Details

To find the volume of HCl that is required to completely neutralize the NaOH solution, we need to use the concept of a neutralization reaction. A neutralization reaction occurs when an acid and a base react to form water and a salt, thus neutralizing each other.

In this particular reaction, the balanced chemical equation is:

HCl + NaOH → NaCl + H₂O

Here, the equation tells us that one mole of HCl reacts with one mole of NaOH. Therefore, the molar ratio of HCl to NaOH is 1:1.

First, let's determine the number of moles of NaOH present in 20 cm³ solution:

Number of moles of NaOH = Concentration (mol/dm³) × Volume (dm³)

= 0.20 mol/dm³ × 20 cm³ × (1 dm³ / 1000 cm³)

= 0.20 × 0.020

= 0.004 moles

Since the reaction is in a 1:1 ratio, the number of moles of HCl required is also 0.004 moles.

Now, let's determine the volume of HCl solution required:

Volume of HCl (dm³) = Number of moles / Concentration

= 0.004 moles / 0.12 mol/dm³

= 0.03333 dm³

Convert this volume from dm³ to cm³:

0.03333 dm³ × 1000 cm³ / dm³ = 33.33 cm³

Therefore, the volume of HCl required is 33.33 cm³.