Question

Consider a solution that is 0.05 M HCl. Your goal is to neutralize 1 L of this solution (i.e. bring the pH to 7). You also have a solution that is 5 M NaOH. What volume of this solution should you add to the HCl solution, to neutralize it? Provide your answer in units of liters (L).

Answer 1

The volume of NaOH required is - 0.01 L

Step-by-step explanation:

At equivalence point ,

Moles of
`HCl` = Moles of NaOH

Considering :-

`Molarity_(HCl)* Volume_(HCl)=Molarity_(NaOH)* Volume_(NaOH)`

Given that:

`Molarity_(NaOH)=5\ M`

`Volume_(NaOH)=?\ L`

`Volume_(HCl)=1\ L`

`Molarity_(HCl)=0.05\ M`

So,

`Molarity_(HCl)* Volume_(HCl)=Molarity_(NaOH)* Volume_(NaOH)`

`0.05* 1=5* Volume_(NaOH)`

`Volume_(NaOH)=(0.05* 1)/(5)=0.01\ L`

The volume of NaOH required is - 0.01 L