Question

What volume of a 0.500 M HCl solution is needed to neutralize each of the following

a. 15..0 mL of a 0.300 M NaOH solution

b. 19.0 mL of a 0.200 M Ba(OH)₂ solution

Answer 1

no of moles of NaOH = molarity of NaOH x volume of NaOH in Liters

= 0.3 x 0.015

= 0.0045 moles

but reaction between NaOH and HCl is

NaOH + HCl ------> NaCl + H2O

so one mole of HCl required one mole of NaOH

so accordingly 0.0045 NaOH moles required 0.0045 moles of HCl

now we know the moles of HCl and molarity of HCl we can calculate the volume of HCl using

volume of HCl = no of moles / molarity

= 0.0045 / 0.5

= 0.009 Liters

= 9 mL of 0.5 HCl

part B

no of moles of Ba(OH)2 = molarity x volume in liters

= 0.2 x 0.019 L

= 0.0038 moles

write the balanced reaction

2HCl + Ba(OH)2 -------> BaCl2 + 2H2O

from this equation it is clear that

for one mole of Ba(OH)2 2 moles of HCl required

so no of moles of HCl = 2 x 0.0038 = 0.0076 moles

volume of HCl = no of moles / Molarity

= 0.0076 / 0.5

= 0.0152 L

= 15.2 mL of 0.5 M HCl