Answer : The volume of 0.10 M NaOH required to neutralize 30 ml of 0.10 M HCl is, 30 ml.
Solution :
According to the neutralization law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of NaOH solution = 0.10 M
[tex]V_1[/tex] = volume of NaOH solution = ?
[tex]M_2[/tex] = molarity of HCl solution = 0.10 M
[tex]V_2[/tex] = volume of HCl solution = 30 ml
Now put all the given values in the above law, we get the volume of NaOH solution.
[tex](0.10M)\times V_1=(0.10M)\times (30ml)[/tex]
[tex]V_1=30ml[/tex]
Therefore, the volume of 0.10 M NaOH required to neutralize 30 ml of 0.10 M HCl is, 30 ml.
