Solution to Chapter 2 Homeworks

Solution to Chapter 2 Homeworks - 1 Chapter 2 Solutions Ch...

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Chapter 2 Solutions Ch 2.5,9,15, 22, 33, 36, 56, 57 2-5 Concentration of sodium bicarbonate Given: 45.000 g of sodium bicarbonate in 1.00 L of water. Solution: a. The resulting concentration of the solution is 45 g/L. Converting this to mg/L (45 g/L)(1000 mg/g) = 45,000 mg/L b. To find the molarity, the molecular weight of NaHCO 3 must be found Na = 22.99 x 1 = 22.99 H = 1.008 x 1 = 1.008 C = 12.01 x 1 = 12.01 3O = 16.00 x 3 = 48.00 Σ = 84.01 g/mol Then, the molarity (45 g/L)(1 mol/84 g) = 0.536 M c. The equivalent weight of NaHCO 3 is its GMW divided by the number of hydrogen ions transferred. In this case n = 1 because Na + is replaced by 1 H. Thus normality (N) is n*M, and in this case n = 1 (0.536 M)(1) = 0.536 M d. Change to units of mg/L as CaCO 3 , using Eqn 2-87 50 mg/L as CaCO 3 45,000 mg/L *(-------------------------) = 26,785 mg/L or 2.68 x 10 4 mg/L as CaCO 3 84 mg/L as NaCO 3 1
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2-9 Concentration of ferric phosphate Given: 2.4 g ferric phosphate added to 1.0 L water, initial phosphate = 1.0 mg/L. Solution: a. Calculate molar concentration of initial phosphate concentration. GMW PO 4 = 94.974.
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