Question

what weight of sodium hydroxide is required to neutralize 100 ml of 0.1N HCl​

Answer 1

Answer:

The weight of sodium hydroxide is required to neutralize 100ml of 0.1N HCl is equal to 0.4g.

Explanation:

The balanced neutralization reaction of NaOH and HCl :-

    NaOH     +     HCl   $\longrightarrow$   NaCl     +   H₂O

In the acid base reaction, there is the formation of salt and water.

In the neutralization reaction of NaOH and HCl, one mole of sodium hydroxide neutralize one mole of hydrochloric acid.

Find the number of moles of 100ml of 0.1N HCl:-

$Normality=\frac{moles}{V \;(in \; L)}$

Moles of HCl = (Normality of the solution of HCl) × (Volume  in L)

Moles of HCl $= (0.1)*(0.1)=0.01 moles$

Therefore, 0.01 moles of sodium hydroxide is required to neutralize 0.01 moles of HCl.

Weight of one mole of NaOH $= 23+16+1=40g$

Then the weight of 0.01mol of NaOH $=40*0.01=0.4g$

Therefore, 0.4g of sodium hydroxide is required to neutralize the 100ml of 0.1N HCl.