What is the Concentration of Sodium Ions in 0.300 m NaCl

Review - Solutions - VI KEY 1) If you dissolve 1.50 mol of aluminum chloride in 2.25 L of water, calculate [ ] of each ion. Answer: [Al³⁺] = 0.667 M [Cl⁻] = 2.00 M 2) If you dissolve 0.240 mol of ammonium sulphate in 320 ml of water, calculate [NH₄⁺](eq). Answer: [NH₄⁺] = 1.50 M 3) If the [Cl⁻] = 0.015 M in 75 ml of water, how many moles of iron (III) chloride were there? Answer: 0.000375 mol FeCl₃ 4) How much water would you need if you started with 1.20 mol of sodium oxalate but wanted a solution with [Na⁺] = 0.48 M? Answer: 5.0 L water 5) In 1.25 L of a 0.50 M nickel (III) sulphate solution, calculate moles of nickel (III) ions. Answer: Ni³⁺ = 1.25 mol SO₄²⁻ = 1.875 mol 6) How many grams of ammonium chloride are present in a 0.30 L beaker of a 0.40 M solution of ammonium chloride? Answer: 6.42 g NH₄Cl 7) How many litres of a 0.250 M potassium chromate solution contains 38.8 g of K₂CrO₄ solution? Answer: 0.799 L K₂CrO₄ KEY 8. A chemist evaporated 25.0 mL of a sodium chloride solution to dryness. He found 0.585 g of NaCl. What was the original concentration of the salt? Answer: 0.585 g NaCl x (1 mol NaCl / 58.44 g NaCl) = 0.0100 mol NaCl 0.0100 mol NaCl / 0.025 L = 0.400 M NaCl 9. If you dissolve 316 g of magnesium bromide in 859 mL of water, calculate [ ]'s of each ion. Answer: 316 g MgBr₂ x (1 mol MgBr₂ / 184.11 g MgBr₂) = 1.716 mol MgBr₂ [Mg²⁺] = 1.716 mol / 0.859 L = 2.00 M [Br⁻] = 4.00 M MgBr₂ → Mg²⁺ + 2Br⁻ 10. 31.1 g of aluminum sulfate is dissolved in 756 mL of water. Calculate [ ] of each ion. Answer: 31.1 g Al₂(SO₄)₃ x (1 mol Al₂(SO₄)₃ / 342.14 g Al₂(SO₄)₃) = 0.0912 mol Al₂(SO₄)₃ [Al³⁺] = 0.240 M [SO₄²⁻] = 0.361 M Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻ 11. 250.0 mL of a 1.2 M Pb(NO₃)₂(aq) solution is diluted to 600.0 mL. Calculate [Pb(NO₃)₂(aq)] and [Pb²⁺] and [NO₃⁻]. Answer: C₁V₁ = C₂V₂ C₂ = (1.2 M)(0.250 L) / (0.600 L) = 0.50 M Pb(NO₃)₂ Pb(NO₃)₂ → Pb²⁺ + 2NO₃⁻ [Pb²⁺] = 0.50 M [NO₃⁻] = 1.0 M 12. 0.300 L of a sodium sulphate solution is diluted to 0.750 L. The diluted concentration of sodium ions is 0.72 M. Calculate the original concentration of the sodium sulphate solution. Answer: C₁V₁ = C₂V₂ C₁ = (0.72 M)(0.750 L) / (0.300 L) = 0.90 M Na₂SO₄ 13. In soda pop, the concentration of dissolved sugar is approximately 0.0125 M. However, in the syrup mix, the sugar concentration is 5.6 M. How many milliliters of the syrup would be needed to make 2.0 L of drinkable soda pop? Answer: C₁V₁ = C₂V₂ V₁ = (0.0125 M)(2.0 L) / (5.6 M) = 0.0045 L = 4.5 mL KEY 14. By accident, 3.57 mL of syrup from the previous question was used instead. Calculate the concentration of the sugar in the 2.0 L soda pop. Answer: C_dil = (C_conc * V_conc) / V_dil C_dil = (5.6 M * 0.00357 L) / (2.0 L) C_dil = 0.010 M 15. A chemist has 46.8 mL of a 0.400 M NaCl solution. How much water should be added to make a 0.250 M solution? Answer: V_dil = (C_conc * V_conc) / C_dil V_dil = (0.400 M * 0.0468 L) / 0.250 M V_dil = 0.0750 L V_water added = 0.0750 L - 0.0468 L = 0.0282 L = 28.2 mL 16. Stephanie adds water to a 125 mL, 3.6 M sodium phosphate solution until she has 1.00 L. Calculate the [ ] of each ion. Answer: Na₃PO₄ → 3Na⁺ + PO₄³⁻ C_dil = (C_conc * V_conc) / V_dil C_dil = (3.6 M * 0.125 L) / (1.00 L) = 0.45 M Na₃PO₄ [Na⁺] = 1.35 M [PO₄³⁻] = 0.45 M 17. 2.0 L of a 0.60 M ferric chloride (iron (III) chloride) solution are mixed with 1.00 L of a 0.90 M barium chloride solution. Calculate the [ ] of each ion. Answer: FeCl₃ → Fe³⁺ + 3Cl⁻ BaCl₂ → Ba²⁺ + 2Cl⁻ [FeCl₃]_dil = (0.60 M * 2.0 L) / (3.0 L) = 0.40 M [BaCl₂]_dil = (0.90 M * 1.00 L) / (3.0 L) = 0.30 M [Fe³⁺] = 0.40 M [Ba²⁺] = 0.30 M [Cl⁻] = 1.2 + 0.60 = 1.8 M 18. 250.0 mL of a 0.30 M potassium sulphate and 450.0 mL of a 0.80 M magnesium chloride solutions are mixed. Calculate the [ ] of each ion. Answer: K₂SO₄ → 2K⁺ + SO₄²⁻ MgCl₂ → Mg²⁺ + 2Cl⁻ [K₂SO₄]_dil = (0.30 M * 0.250 L) / (0.700 L) = 0.10714 M [MgCl₂]_dil = (0.80 M * 0.450 L) / (0.700 L) = 0.51428 M [K⁺] = 0.21 M [Mg²⁺] = 0.51 M [SO₄²⁻] = 0.11 M [Cl⁻] = 1.0 M KEY 19. 20.83 g of BaCl₂ is dissolved in 75.0 mL of 2.00 M NaCl. Calculate the concentrations of all ions. Answer: 20.83 g BaCl₂ x (1 mol BaCl₂ / 208.23 g BaCl₂) x (1 / 0.075 L) = 1.3334 M BaCl₂ BaCl₂ → Ba²⁺ + 2Cl⁻ [Ba²⁺] = 1.33 M [Cl⁻] = 2.00 + 2.66 = 4.66 M [Na⁺] = 2.00 M 20. 71.0 g of aluminum nitrate and 53.4 g of magnesium nitrate are dissolved in 800.0 mL of water. Calculate [Al(NO₃)₃], [Mg(NO₃)₂], and [NO₃⁻]. Answer: 71.0 g Al(NO₃)₃ x (1 mol Al(NO₃)₃ / 212.99 g Al(NO₃)₃) x (1 / 0.800 L) = 0.4166 M Al(NO₃)₃ 53.4 g Mg(NO₃)₂ x (1 mol Mg(NO₃)₂ / 148.33 g Mg(NO₃)₂) x (1 / 0.800 L) = 0.4500 M Mg(NO₃)₂ Al(NO₃)₃ → Al³⁺ + 3NO₃⁻ [Al(NO₃)₃] = 0.417 M [NO₃⁻] = 1.25 + 0.900 = 2.15 M Mg(NO₃)₂ → Mg²⁺ + 2NO₃⁻ [Mg(NO₃)₂] = 0.450 M 21. 123.0 g of sodium acetate and 0.48 mol of strontium acetate are dissolved in 496 mL of water. Calculate the concentration of acetate. Answer: 0.48 mol Sr(CH₃COO)₂ x (1 / 0.496 L) = 0.96774 M Sr(CH₃COO)₂ 123.0 g NaCH₃COO x (1 mol NaCH₃COO / 82.03 g NaCH₃COO) x (1 / 0.496 L) = 3.0227 M NaCH₃COO NaCH₃COO → Na⁺ + CH₃COO⁻ Sr(CH₃COO)₂ → Sr²⁺ + 2CH₃COO⁻ [CH₃COO⁻] = 3.02 + 1.94 = 4.96 M 22. 40.3 g of ferric nitrate and 97.38 g of ferric chloride are dissolved in 0.25 L of water. Calculate the concentration of the ferric iron ion (Fe³⁺). Answer: 40.3 g Fe(NO₃)₃ x (1 mol Fe(NO₃)₃ / 241.86 g Fe(NO₃)₃) x (1 / 0.25 L) = 0.6664 M Fe(NO₃)₃ 97.38 g FeCl₃ x (1 mol FeCl₃ / 162.20 g FeCl₃) x (1 / 0.25 L) = 2.4014 M FeCl₃ Fe(NO₃)₃ → Fe³⁺ + 3NO₃⁻ FeCl₃ → Fe³⁺ + 3Cl⁻ [Fe³⁺] = 0.66 + 2.40 = 3.0 M Ca(NO₃)₂ → Ca²⁺ + 2NO₃⁻ [Ca(NO₃)₂]dil = (0.80 M)(0.150 L) / (0.800 L) = 0.150 M [Ca²⁺] = 0.150 M [NO₃⁻] = 0.30 + 0.72 = 1.02 M KNO₃ → K⁺ + NO₃⁻ [KNO₃]dil = 0.975 M [K⁺] = 0.975 M [NO₃⁻] = 0.975 + 0.30 = 1.3 M 24. 25 mL of a 4.4 M magnesium sulphate solution are mixed with 75.0 mL of a 1.25 M magnesium phosphate solution. Calculate the concentrations of all ions of the dilute solution. Answer: MgSO₄ → Mg²⁺ + SO₄²⁻ [MgSO₄]dil = (4.4 M)(0.025 L) / (0.100 L) = 1.1 M [Mg²⁺] = 1.1 + 2.8 = 3.9 M [SO₄²⁻] = 1.1 M Mg₃(PO₄)₂ → 3Mg²⁺ + 2PO₄³⁻ [Mg₃(PO₄)₂]dil = (1.25 M)(0.075 L) / (0.100 L) = 0.9375 M [PO₄³⁻] = 1.9 M 25. 0.200 L of a 0.050 M AlBr₃ solution is added to 50.0 mL of a 0.40 M KOH solution. a.) Will a precipitate form? If so, what is it? Answer: Yes, Al(OH)₃ b.) Write a net ionic equation. Answer: Al³⁺ + 3OH⁻ → Al(OH)₃(s) c.) Calculate the mass of the precipitate, if a precipitate formed. Answer: 0.200 L AlBr₃ x (0.050 mol AlBr₃ / 1 L) x (1 mol Al(OH)₃ / 1 mol AlBr₃) = 0.0010 mol Al(OH)₃ 0.0010 mol Al(OH)₃ x (78 g Al(OH)₃ / 1 mol Al(OH)₃) = 0.078 g Al(OH)₃ d.) If 0.65 g of precipitate are recovered, calculate the % yield. Answer: percent yield = (product obtained / product expected) x 100% percent yield = (0.65 g / 0.078 g) x 100% = 125%