(a) What volume of \(0.115 \mathrm{M} \mathrm{HClO}_{4}\) solution is needed to neutralize \(50.00 \mathrm{~mL}\) of \(0.0875 \mathrm{M} \mathrm{NaOH}\) ? (b) What volume of \(0.128 \mathrm{M} \mathrm{HCl}\) is needed to neutralize \(2.87 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\) (c) If \(25.8 \mathrm{~mL}\) of \(\mathrm{AgNO}_{3}\) is needed to precipitate all the \(\mathrm{Cl}^{-}\) ions in a \(785-\mathrm{mg}\) sample of \(\mathrm{KCl}\) (forming \(\mathrm{AgCl}\) ), what is the molarity of the \(\mathrm{AgNO}_{3}\) solution? (d) If \(45.3 \mathrm{~mL}\) of \(0.108 \mathrm{M} \mathrm{HCl}\) solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

Step by step solution

01

Identify the neutralization reaction

The neutralization reaction between HClO₄ and NaOH is: \( HClO_4 + NaOH \rightarrow NaClO_4 + H_2O \)

02

Calculate the moles of NaOH

Moles of NaOH = Molarity × Volume in liters Moles of NaOH = 0.0875 × (50.00/1000) = 0.004375 moles

03

Calculate the moles of HClO₄

From the balanced equation, the stoichiometry is 1:1, so moles of HClO₄ = moles of NaOH = 0.004375 moles

04

Calculate the volume of HClO₄

Volume of HClO₄ (L) = Moles / Molarity Volume of HClO₄ = 0.004375 / 0.115 = 0.038044 L = 38.04 mL (Approximately) Problem (b)

05

Calculate the moles of Mg(OH)₂

Moles of Mg(OH)₂ = Mass / Molar mass Moles of Mg(OH)₂ = 2.87 / (24.305 + 34.02) = 0.0354 moles

06

Identify the neutralization reaction

The neutralization reaction between HCl and Mg(OH)₂ is: \( Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O \)

07

Calculate the moles of HCl

From the balanced equation, the stoichiometry is 2:1, so moles of HCl = 2 × moles of Mg(OH)₂ = 2 × 0.0354 = 0.0708 moles

08

Calculate the volume of HCl

Volume of HCl (L) = Moles / Molarity Volume of HCl = 0.0708 / 0.128 = 0.553 L = 553 mL (Approximately) Problem (c)

09

Calculate moles of KCl

Moles of KCl = Mass / Molar mass Moles of KCl = 0.785 / (39.098 + 35.453) = 0.01033 moles

10

Calculate moles of Cl⁻ ions

Moles of Cl⁻ ions = moles of KCl = 0.01033 moles

11

Calculate the molarity of AgNO₃

Molarity of AgNO₃ = Moles of Cl⁻ ions / Volume of AgNO₃ (L) Molarity of AgNO₃ = 0.01033 / (25.8 / 1000) = 0.4004 M Problem (d)

12

Calculate the moles of HCl

Moles of HCl = Molarity × Volume in liters Moles of HCl = 0.108 × (45.3 / 1000) = 0.004886 moles

13

Identify the neutralization reaction

The neutralization reaction between HCl and KOH is: \( HCl + KOH \rightarrow KCl + H_2O \)

14

Calculate the moles of KOH

From the balanced equation, the stoichiometry is 1:1, so moles of KOH = moles of HCl = 0.004886 moles

15

Calculate the mass of KOH

Mass of KOH = Moles × Molar mass Mass of KOH = 0.004886 × (39.098 + 15.999) = 0.269 g (Approximately)

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