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

Answer:

The volume of NaOH required is - 0.01 L

Explanation:

At equivalence point ,

Moles of $HCl$ = Moles of NaOH

Considering :-

$Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}$

Given  that:

$Molarity_{NaOH}=5\ M$

$Volume_{NaOH}=?\ L$

$Volume_{HCl}=1\ L$

$Molarity_{HCl}=0.05\ M$

So,  

$Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}$

$0.05\times 1=5\times Volume_{NaOH}$

$Volume_{NaOH}=\frac{0.05\times 1}{5}=0.01\ L$

The volume of NaOH required is - 0.01 L