Question

What volume in milliliters of 0.0180 M Ba(OH)₂ is required to neutralize 55.0 mL of 0.0500 M HCl?

Answer 1

The volume in milliliters of 0.0180 M Ba(OH)₂ that is required to neutralize 55.0 mL of 0.0500 M HCl is 152.78mL.

HOW TO CALCULATE VOLUME?

The volume of a solution can be calculated by using the following expression:

C1V1 = C2V2

Where;

• C1 = initial concentration
• C2 = final concentration
• V1 = initial volume
• V2 = final volume

The volume can be calculated as follows;

0.0180 × V1 = 55 × 0.05

0.0180V1 = 2.75

V1 = 2.75 ÷ 0.0180

V1 = 152.78mL

Therefore, the volume in milliliters of 0.0180 M Ba(OH)₂ that is required to neutralize 55.0 mL of 0.0500 M HCl is 152.78mL.

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