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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).

Respuesta :

Answer:

The volume of NaOH required is - 0.01 L

Explanation:

At equivalence point ,

Moles of [tex]HCl[/tex] = Moles of NaOH

Considering :-

[tex]Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}[/tex]

Given  that:

[tex]Molarity_{NaOH}=5\ M[/tex]

[tex]Volume_{NaOH}=?\ L[/tex]

[tex]Volume_{HCl}=1\ L[/tex]

[tex]Molarity_{HCl}=0.05\ M[/tex]

So,  

[tex]Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}[/tex]

[tex]0.05\times 1=5\times Volume_{NaOH}[/tex]

[tex]Volume_{NaOH}=\frac{0.05\times 1}{5}=0.01\ L[/tex]

The volume of NaOH required is - 0.01 L

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