How many milliliters of 0.100 M Ba(OH)₂ are required to neutralize 20.0 mL of 0.250 M HCl?
A)100. mL
B)50.0 mL
C)25.0 mL
D)0.250 mL
E)0.50 mL

Question

Quizplus · Accepted Answer

We can use the equation for an acid-base reaction to find the amount of Ba(OH)₂ required to neutralize the HCl:

HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

The balanced equation tells us that 1 mole of HCl reacts with 1 mole of Ba(OH)₂. We can use the molarity and volume of the HCl to find the number of moles of HCl:

0.250 M HCl × 0.0200 L HCl = 0.00500 moles HCl

Since we need 1 mole of Ba(OH)₂ for every mole of HCl, we need 0.00500 moles of Ba(OH)₂. We can use the molarity of the Ba(OH)₂ to find the volume needed:

0.100 M Ba(OH)₂ = 0.100 moles Ba(OH)₂ / 1 L solution

0.00500 moles Ba(OH)₂ × (1 L solution / 0.100 moles Ba(OH)₂) = 0.0500 L = 50.0 mL

Therefore, the answer is C) 25.0 mL.

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