Question

Determine the volume of 0.150 M NaOH solution required to neutralize each sample of hydrochloric acid....

Determine the volume of 0.150 M NaOH solution required to neutralize each sample of hydrochloric acid. The neutralization reaction is: NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) 25 mL of a 0.150 M HCl solution 55 mL of a 0.055 M HCl solution 175 mL of a 0.885 M HCl solution Express your answers, separated by commas, in liters.

Answer – We are given, [NaOH] = 0.150 M

Reaction - NaOH(aq) + HCl(aq) -----> H2O(l) + NaCl(aq)

So mole ration is for all is 1:1.

a)25 mL of a 0.150 M HCl solution

We need to calculate the moles of HCl

Moles of HCl = 0.150 M * 0.025 mL

= 0.00375 moles

Moles of HCl = moles of NaOH = 0.00375 moles

So, volume of NaOH required = 0.00375 moles / 0.150 M = 0.025 L

b) 55 mL of a 0.055 M HCl solution

We need to calculate the moles of HCl

Moles of HCl = 0.055 M * 0.055 mL

= 0.00303 moles

Moles of HCl = moles of NaOH = 0.00303 moles

So, volume of NaOH required = 0.00303 moles / 0.150 M = 0.0202 L

c) 175 mL of a 0.885 M HCl

We need to calculate the moles of HCl

Moles of HCl = 0.885 M * 0.175 mL

= 0.155 moles

Moles of HCl = moles of NaOH = 0.155 moles

So, volume of NaOH required = 0.155 moles / 0.150 M = 1.03 L

So answers are = 0.025 L , 0.0202 L, 1.03 L