Describe how you would prepare each of the following aqueous solutions: (a) 1.50 \(\mathrm{L}\) of 0.110 \(\mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution, starting with solid \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ;\) (b) 225 \(\mathrm{g}\) of a solution that is 0.65 \(\mathrm{m}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) starting with the solid solute; ( c ) 1.20 L of a solution that is 15.0\(\% \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass (the density of the solution is 1.16 \(\mathrm{g} / \mathrm{mL}\) , starting with solid solute; (\boldsymbol{d} ) ~ a ~ 0.50 \(\mathrm{M}\) solution of HCl that would just neutralize 5.5 \(\mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with 6.0 \(\mathrm{M} \mathrm{HCl}\) .

Step by step solution

01

To find the moles of (NH₄)₂SO₄ needed, we can use the formula: moles = Molarity (M) × Volume (L) moles = 0.110 M × 1.50 L = 0.165 mol of (NH₄)₂SO₄ Step 2: Convert moles to grams

To convert moles to grams, we use the formula: mass = moles × molar mass The molar mass of (NH₄)₂SO₄ = (2 × (1.0079 × 2 + 14.0067)) + (32.065 + 4 × 15.9994) = 132.14 g/mol mass = 0.165 mol × 132.14 g/mol = 21.80 g of (NH₄)₂SO₄ Step 3: Mix the solute and solvent

02

Measure 21.80 g of solid (NH₄)₂SO₄ and dissolve it in a 1.50 L volumetric flask with distilled water up to the mark. #b) Preparing 225 g of a 0.65 m Na₂CO₃ solution from solid Na₂CO₃# Step 1: Calculate moles of solute

To find the moles of Na₂CO₃, we can use the formula: moles = molality (m) × mass of solvent (kg) Assuming the mass of the solution (mass_solution) is 225 g. Let the mass of the solute (mass_solute) be x grams, and the mass of the solvent (water) is (225−x) grams. moles = 0.65 m × (225−x) g / 1000 = (0.65 × (225−x)/1000) mol of Na₂CO₃ Step 2: Convert moles to grams

03

The molar mass of Na₂CO₃ = (2 × 22.98977) + 12.0107 + (3 × 15.9994) = 105.99 g/mol mass_solute = x = moles × molar mass = (0.65 × (225−x)/1000) × 105.99 g/mol Step 3: Solve for x (mass of Na₂CO₃)

Solve the equation for x: x = (0.65 × (225−x)/1000) × 105.99 x = 14.29 g of Na₂CO₃ Step 4: Mix the solute and solvent

04

Measure 14.29 g of solid Na₂CO₃ and dissolve it in a container with enough distilled water to make a total mass of 225 g of the solution. #c) Preparing 1.20 L of a solution that is 15.0% Pb(NO₃)₂ by mass; density = 1.16 g/mL# Step 1: Calculate mass of the solution

To find the mass of the solution (mass_solution), we use the formula: mass_solution = volume × density mass_solution = 1.20 L × 1000 mL/L × 1.16 g/mL = 1392 g of solution Step 2: Calculate mass of Pb(NO₃)₂

05

To find the mass of Pb(NO₃)₂, we can use the formula: mass_solute = (% by mass × mass_solution) / 100 mass_solute = (15.0 × 1392) / 100 = 208.8 g of Pb(NO₃)₂ Step 3: Mix the solute and solvent

Measure 208.8 g of solid Pb(NO₃)₂ and dissolve it in a 1.20 L volumetric flask with distilled water up to the mark. #d) Preparing 0.50 M HCl solution that would just neutralize 5.5 g of Ba(OH)₂, starting with 6.0 M HCl# Step 1: Calculate moles of Ba(OH)₂

06

First, we need to calculate the moles of Ba(OH)₂ using formula: moles = mass / molar mass The molar mass of Ba(OH)₂ = 137.327 + 2 × (15.9994 + 1.0079 × 2) = 171.36 g/mol moles of Ba(OH)₂ = 5.5 g / 171.36 g/mol = 0.0321 mol Step 2: Calculate moles of HCl

For the neutralization reaction: Ba(OH)₂ + 2 HCl → BaCl₂ + 2 H₂O The stoichiometry tells us that 1 mol of Ba(OH)₂ reacts with 2 mol of HCl. moles of HCl = 2 × 0.0321 mol = 0.0642 mol Step 3: Calculate volume of 6.0 M HCl

07

To find the volume of 6.0 M HCl required, we can use the formula: volume = moles / molarity volume of 6.0 M HCl: 0.0642 mol / 6.0 M = 0.0107 L or 10.7 mL Step 4: Dilute the HCl solution

Dilute 10.7 mL of 6.0 M HCl with distilled water in a volumetric flask to obtain a final volume of 0.0642 mol ÷ 0.50 M ≈ 0.128 L (128 mL) (0.50 M HCl solution that would just neutralize 5.5 g of Ba(OH)₂).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!