Question

What volume of 0.108 mol L^-1 NaOH solution is needed to just completely neutralize a 0.459 mol L^-1 solution of CH₃COOH with volume 15.1 mL?

The answer should be expressed in mL.

Answer 1

1 mole CH3COOH dissociates to give 1 mole of CH3COO- and 1 mole of H+ to neutralize 1 mole of CH3COOH 1 mole of NaOH is required because NaOH dissociates to give 1 mole of  Na+ and OH-, so 1 mole of NaOH is required to neutralize 1 mole of CH3COOH

here

concnetration of CH3COOH is 0.459 mol/ L in 1 L there are 0.459 moles of CH3COOH, that is in 1000 mL there are 0.459 moles of CH3COOH

so number of moles present in 1 mL= 0.459 / 1000

so number of moles present in 15.1 mL= 15.1*0.459/1000

= 0.0069309 moles

so we need  0.0069309 moles of NaOH, concentration of NaOH solution is 0.108 mol/L,

there are 0.108 moles in 1 L that is in 1000 mL

so number of moles in 1 mL= 0.108/1000

= 0.000108 moles

so volume of NaOH required to neutralize CH3COOH= number of moles of CH3COOH present in 15.1 mL/ number of moles present in 1 mL of NaOH

= 0.0069309 moles/ 0.000108 moles

=64.175 mL