Question

Name Pre Lab Questions Date A. Calculate mass of Cuso, 5H.0 to prepare 25mL of 0.100M solution. (5 points each, 25 points total) 0.1com L 25 ml. 0.100 ML 100 mL B. Calculate the volume of a 0.125M solution of NaOH that is needed to make 50mL of a 0.075M solution. C. Provide the concentration of aluminum ion and the concentration of acetate ion in a 0.33M solution of aluminum acetate. Aluminum ion, M Acetate ion, M D. A SOL solution of 0.150M NaCl is added to a 40mL. solution of 0.170M AgNO, Write the balanced chemical reaction. including state symbols. Calculate the concentrations (M) of the ions after the reaction Sodium ion, M Nitrate ion, M C silver ion, M hloride ion, M Chemical reaction: E How many drops of water must you add to get O.IM Na PO, from a solution that has 1 drop of 0.5M NA PO 2

Answer 1

(A.) Concentration CuSO₄.5H₂O = 0.100 M

volume of solution = 25 mL = 0.025 L

moles CuSO₄.5H₂O needed = (Concentration CuSO₄.5H₂O) * (volume of solution in Liter)

moles CuSO₄.5H₂O needed = (0.100 M) * (0.025 L)

moles CuSO₄.5H₂O needed = 0.0025 mol

mass CuSO₄.5H₂O needed = (moles CuSO₄.5H₂O needed) * (molar mass CuSO₄.5H₂O)

mass CuSO₄.5H₂O needed = (0.0025 mol) * (249.7 g/mol)

mass CuSO₄.5H₂O needed = 0.624 g

(B.) According to dilution law,

C1 * V1 = C2 * V2

where C1 = initial concentration of NaOH = 0.125 M

V1 = volume of original solution required

C2 = final concentration of NaOH = 0.075 M

V2 = final volume of solution = 50 mL

V1 = V2 * (C2 / C1)

V1 = (50 mL) * (0.075 M / 0.125 M)

V1 = (50 mL) * (0.6)

V1 = 30 mL

volume of 0.125 M solution of NaOH needed = 30 mL

(C.) aluminum acetate is Al(CH₃COO)₃

Al(CH₃COO)₃ (aq) $\rightarrow$ Al³⁺ (aq) + 3 CH₃COO^- (aq)

concentration of aluminum ion = concentration of aluminum acetate

concentration of aluminum ion = 0.33 M

concentration of acetate ion = 3 * (concentration of aluminum acetate)

concentration of acetate ion = 3 * (0.33 M)

concentration of acetate ion = 0.99 M

(D.) The balanced chemical reaction is :

NaCl (aq) + AgNO₃ (aq) $\rightarrow$ AgCl (s) + NaNO₃ (aq)

concentration NaCl = 0.150 M

volume NaCl solution = 50 mL

moles NaCl = (concentration NaCl) * (volume NaCl solution)

moles NaCl = (0.150 M) * (50 mL)

moles NaCl = 7.5 mmol

initial moles Na⁺ = moles NaCl = 7.5 mmol

initial moles Cl^- = moles NaCl = 7.5 mmol

Similarly, moles AgNO₃ = 6.8 mmol

initial moles Ag⁺ = moles AgNO₃ = 6.8 mmol

initial moles NO₃^- = moles AgNO₃ = 6.8 mmol

Since less moles of AgNO₃ are present, therefore AgNO₃ is the limiting reactant

moles AgCl formed = moles AgNO₃ consumed

moles AgCl formed = 6.8 mmol

moles Ag⁺ left = 0

concentration of silver ion after reaction = 0

moles Cl^- left = (initial moles Cl^-) - (moles Cl^- consumed in AgCl)

moles Cl^- left = (7.5 mmol) - (6.8 mmol)

moles Cl^- left = 0.7 mmol

Total volume = (50 mL) + (40 mL) = 90 mL

concentration of Cl^- ion after reaction = (moles Cl^- left) / (total volume)

concentration of Cl^- ion after reaction = (0.7 mmol) / (90 mL)

concentration of Cl^- ion after reaction = 0.0078 M

concentration of Na⁺ ion after reaction = (moles Na⁺) / (total volume)

concentration of Na⁺ ion after reaction = (7.5 mmol) / (90 mL)

concentration of Na⁺ ion after reaction = 0.083 M

concentration of NO₃^- ion after reaction = (moles NO₃^-) / (total volume)

concentration of NO₃^- ion after reaction = (6.8 mmol) / (90 mL)

concentration of NO₃^- ion after reaction = 0.076 M

(E.) According to dilution law,

C1 * V1 = C2 * V2

where C1 = initial concentration of Na₃PO₄ = 0.5 M

V1 = volume of original solution = 1 drop

C2 = final concentration of NaOH = 0.1 M

V2 = final volume of solution

V2 = V1 * (C1 / C2)

V2 = (1 drop) * (0.5 M / 0.1 M)

V2 = (1 drop) * (5)

V2 = 5 drops

drops of water added = (final volume) - (initial volume)

drops of water added = (5 drops) - (1 drop)

drops of water added = 4 drops