Answer: The volume of 0.10 M NaOH required to neutralize 30 ml of 0.10 M HCl is, 30 ml.
Explanation:
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock [tex]HCl[/tex] solution = 12.0 M
[tex]V_1[/tex] = volume of stock [tex]HCl[/tex]solution = 5.0 ml
[tex]M_2[/tex] = molarity of dilute [tex]HCl[/tex] solution = 0.2 M
[tex]V_2[/tex] = volume of dilute [tex]HCl[/tex] solution = ?
Putting in the values we get:
[tex](12.0M)\times 5.0=0.2\times V_2[/tex]
[tex]V_2=300ml[/tex]
Therefore, 300 ml of 0.2M HCl can be made from 5.0mL of a 12.0M [tex]Hcl[/tex] solution.
