Question

A solution of NaOH is 4 g per litre. What volume of HCl gas at STP will neutralise, 50 ml of the alkali solution ?? 
(Please explain with proper Method)

Answer 1

Answer : The volume of HCl gas at STP will be, 0.112 L or 112 ml

Explanation :

Molar mass of NaOH = 40 g/mole

First we have to calculate the mass of NaOH.

As, 1 L of solution contains 4 g of NaOH

So, 0.05 L of solution contains $4\times 0.05=0.2g$ of NaOH

The mass of NaOH = 0.2 g

Now we have to calculate the moles of NaOH.

$\text{Moles of NaOH}=\frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}}=\frac{0.2g}{40g/mole}=0.005mole$

The moles of NaOH = 0.005 mole

Now we have to calculate the moles of HCl.

The balanced chemical reaction will be,

$NaOH+HCl\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that

As, 1 mole of NaOH neutralizes 1 mole of HCl

So, 0.005 mole of NaOH neutralizes 0.005 mole of HCl

The moles of HCl = 0.005 mole

Now we have to calculate the volume of HCl.

As we know that at STP,

1 mole of HCl contains 22.4 L volume of HCl

So, 0.005 mole of HCl contains $22.4\times 0.005=0.112L=112ml$ volume of HCl

Therefore, the volume of HCl gas at STP will be, 0.112 L or 112 ml