Question

What volume in milliliters of 0.0130 M Ca(OH)₂ is required to neutralize 75.0 mL of 0.0300 M HCl?

Answer 1

The volume in milliliters of calcium hydroxide solution required to neutralize 75.0 mL of 0.0300 M HCl is 173.1mL

HOW TO CALCULATE MOLARITY:

• The molarity of calcium chloride solution can be calculated by using the following formula:

C1V1 = C2V2

Where;

1. C1 = concentration of calcium hydroxide
2. V1 = volume of calcium hydroxide
3. C2 = concentration of hydrogen chloride
4. V2 = volume of hydrogen chloride

According to this question,

1. C1 = 0.0130M
2. C2 = 0.0300 M
3. V1 = ?
4. V2 = 75.0 mL

• 0.0130 × V1 = 0.0300 × 75

• 0.0130V1 = 2.25

• V1 = 2.25 ÷ 0.0130

• V1 = 173.1mL

• Therefore, the volume in milliliters of calcium hydroxide solution required to neutralize 75.0 mL of 0.0300 M HCl is 173.1mL.

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