Question

How many milliliters of 0.100 M Ba(OH)2 are required to neutralize 20.0 mL of 0.250 M HCl?

Answer 1

Ba(OH)₂ + 2HCl = BaCl₂ + 2H₂O

n(HCl)=c(HCl)v(HCl)

n{Ba(OH)₂)}=c{Ba(OH)₂}v{Ba(OH)₂}=2n(HCl)=2c(HCl)v(HCl)

v{Ba(OH)₂}=2c(HCl)v(HCl)/c{Ba(OH)₂}

v{Ba(OH)₂}=2*0.250*0.020/0.100=0.100 L = 100 mL

Answer 2

Answer:

25.0 mL of 0.100 M Ba(OH)₂ are required.

Explanation:

The reaction that takes place is:

Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O

The formula for concentration is:

C = n / V

• With the volume and concentration of the HCl solution, we can calculate the moles of HCl, keeping in mind that 20.0 mL = 0.020 L:

0.020 L * 0.250 M = 0.005 mol HCl

Then we convert them to moles of Ba(OH)₂:

0.005 mol HCl * $\frac{1molBa(OH)_{2}}{2molHCl}$ = 0.0025 mol  Ba(OH)₂

• Finally, with the moles of Ba(OH)₂ and the concentration we can calculate the volume:

0.0025 mol Ba(OH)₂ / 0.100 M = 0.025 L = 25 mL