Problem 7.66 How many milliliters of 9.00 M KOH must be added to neutralize the

Question

Problem 7.66

How many milliliters of 9.00 M KOH must be added to neutralize the following solutions?

Part A

a mixture of 0.240 M LiOH (25.0 mL) and 0.200 M HBr (75.0 mL)

Express your answer with the appropriate units.

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Part B

a mixture of 0.300 M HCl (45.0 mL) and 0.250 M NaOH (10.0 mL)

Express your answer with the appropriate units.

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Problem 7.66

How many milliliters of 9.00 M KOH must be added to neutralize the following solutions?

Part A

a mixture of 0.240 M LiOH (25.0 mL) and 0.200 M HBr (75.0 mL)

Express your answer with the appropriate units.

VKOH =

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Incorrect; Try Again; 4 attempts remaining

Part B

a mixture of 0.300 M HCl (45.0 mL) and 0.250 M NaOH (10.0 mL)

Express your answer with the appropriate units.

VKOH =

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Explanation / Answer

Balanced equation:
LiOH + HBr = LiBr + H2O

Part A

Moles of LiOH = 25 x 0.24 /1000 = 0.006 Moles

Moles of HBr = 75 x 0.2 /1000 = 0.015 Moles

Moles of HBr after reacting with LiOH = 0.015 -0.006 = 0.009 Moles

Moles of KOH to be needed = 0.009 Moles

Volume of KOH = 0.009 x 1000 /9 = 1 ml

Hence 1 ml of KOH is needed

Part B

Moles of HCl = 0.3 x 45 /1000 = 0.0135 Moles

Moles of NaOH = 0.25 x 10 / 1000 = 0.0025 Moles

Moles of HCl after reacting with NaOH = 0.0135 - 0.0025 = 0.011 Moles

Moles of KOH = 0.011 Moles

Volume of KOH = 0.011 x 1000 /9 = 1.2 ml

1.2 ml of KOH is needed

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