Question

What volume (in mL) of a 0.0557 M Sr(OH)2 solution is needed to neutralize 36.4 mL of a 0.0750 M HN03 solution

Answer 1

The volume (in mL) of a 0.0557 M Sr(OH)2 solution is needed to neutralize 36.4 mL of a 0.0750 M HN03 solution is 49mL.

HOW TO CALCULATE VOLUME:

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

C1V1 = C2V2

Where;

• C1 = base concentration (M)
• C2 = acid concentration (M)
• V1 = base volume (mL)
• V2 = acid volume (mL)

According to this question;

• C1 = 0.0557 M
• C2 = 0.0750 M
• V1 = ?
• V2 = 36.4 mL

0.0557 × V1 = 0.0750 × 36.4

0.0557V1 = 2.73

V1 = 2.73 ÷ 0.0557

V1 = 49.0

Therefore, the volume (in mL) of a 0.0557 M Sr(OH)2 solution is needed to neutralize 36.4 mL of a 0.0750 M HN03 solution is 49mL.

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