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

To neutralize 20.0 mL of 0.250 M HCl, 25.0 mL of 0.100 M Ba(OH)2 is required, as determined by a stoichiometry calculation based on the balanced chemical equation for the reaction.

Step-by-step explanation:

To determine how many milliliters of 0.100 M Ba(OH)2 are required to neutralize 20.0 mL of 0.250 M HCl, we need to use a stoichiometry calculation based on the balanced chemical equation for the reaction:

Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O

From the equation, we see that one mole of Ba(OH)2 reacts with two moles of HCl. First, calculate the number of moles of HCl that are present in the 20.0 mL solution:

0.250 mol/L * 0.020 L = 0.005 moles of HCl

Since it takes 1 mole of Ba(OH)2 to neutralize 2 moles of HCl, we only need 0.0025 moles of Ba(OH)2:

0.005 moles HCl * (1 mole Ba(OH)2/2 moles HCl) = 0.0025 moles Ba(OH)2

Finally, calculate the volume of 0.100 M Ba(OH)2 needed to provide 0.0025 moles:

0.0025 moles / 0.100 M = 0.025 L or 25.0 mL

Therefore, 25.0 mL of 0.100 M Ba(OH)2 are required to neutralize 20.0 mL of 0.250 M HCl.

Answer 2

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

Step-by-step 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 *
`(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