50cm3 of 0.2mol/dm3 hydrochloric acid was reacted with 0.6g of magnesium. The resulting solution was titrated against sodium hydroxide of concentration 1.67mol/dm3.Calculate the volume of alkaline needed.

0.06 liter

Explanation:

First, let us look at the equation of the first reaction:

2HCl + Mg > MgCl2 + H2

2 moles of HCl requires 1 mole of Mg for complete reaction.

Mole of HCl available = molarity x volume

                                    0.2 x 50/1000 = 0.1 mole

Mole of Mg available = mass/molar mass

                                    0.6/24.3 = 0.02 mole.

Thus, HCl is in excess by 0.1 mole.

Now, let us look at the second reaction:

HCl + NaOH > NaCl + H2O

mole ratio of HCl to NaOH is 1:1

Remember that 0.1 mole HCl is left from the first reaction. Thus, 0.1 mole of the alkaline would also be needed.

Hence, volume of NaOH needed = mole/molarity

                        0.1/1.67 = 0.06 dm3

The volume of alkaline needed would, therefore, be 0.06 liter.

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