Answer:
d)50.0 mL
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}=3.00\ M[/tex]
[tex]Volume_{NaOH}=100\ mL[/tex]
[tex]Volume_{HCl}=?\ mL[/tex]
[tex]Molarity_{HCl}=3.00\ M[/tex]
So,
[tex]Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}[/tex]
[tex]3.00\times Volume_{HCl}=3.00\times 100[/tex]
[tex]Volume_{HCl}=\frac{3.00\times 100}{3.00}\ mL=100\ mL[/tex]
The volume of HCl at equivalence point is - 100 mL
More addition of the acid forms a acidic solution. Thus, d)50.0 mL is the correct option because all other options are greater or equal to 100 which leads to acidic or neutral solution respectively.
