Question

# A) What volume of 0.150 M HClO4 solution is needed to neutralize 59.00 mL of 8.80×10−2...

A) What volume of 0.150 M HClO4 solution is needed to neutralize 59.00 mL of 8.80×10−2 M NaOH? B) What volume of 0.130 M HCl is needed to neutralize 2.87 g of Mg(OH)2? C) If 26.4 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.755-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution? D) If 45.8 mL of 0.112 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

A)

Moles of NaOH = Moles of HClO4

Moles of NaOH = 0.059 x 8.8 x 10^-2 = 0.005192 moles

=> Moles of HClO4 required = 0.005192

Molarity = Moles / Volume

=> 0.15 = 0.005192 / Volume

=> Volume = 0.03461 L = 34.61 mL

B)

Mass of Mg(OH)2 = 2.87 g

=> Moles of Mg(OH)2 = 2.87 / 58.32 = 0.0492 moles

=> Moles of HCl required = 0.0492 x 2 = 0.0984 moles

=> Volume of HCl = 0.0984 / 0.13 = 0.7571 L = 757.1 mL

C)

Mass of KCl = 0.755 mg = 7.55 x 10^-4 g

=> Moles of KCl = 7.55 x 10^-4 / 74.55 = 1.013 x 10^-5 moles

=> Moles of AgNO3 required = 1.013 x 10^-5 moles

=> Molarity = 1.013 x 10^-5 / 0.0264 = 3.84 x 10^-4 M

D)

Moles of HCl = Mole sof KOH = 0.0458 x 0.112 = 5.13 x 10^-3 moles

=> Mass of KOH = 5.13 x 10^-3 x 56.11 = 0.288 g

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