Respuesta :
So,
The balanced reaction for this situation is:
[tex]NaOH_{(aq)}+HCl_{(aq)}\to NaCl_{(s)}+H_2O_{(l)}[/tex]The first thing we're going to do is to determine the # of moles of HCl from its volume and molarity: (Remember that 5mL are 0.005 liters)
[tex]\begin{gathered} M=\frac{n}{V}\to n=M\cdot V \\ \\ n=0.0500M\cdot0.005L \\ n=0.00025molesHCl \end{gathered}[/tex]Now, we're going to convert these moles to moles of NaOH using stoichiometry:
[tex]0.00025molesHCl\cdot\frac{1molNaOH}{1molHCl}=0.00025molesNaOH[/tex]And then, use the calculated moles of NaOH and its molarity (0.100 M) to figure out the volume needed.
[tex]V=\frac{n}{M}=\frac{0.00025molesNaOH}{0.1}=0.0025L[/tex]If we pass this answer to mL:
[tex]0.0025L=2.5mL[/tex]Therefore, we need 2.5mL of NaOH to neutralize the solution.
