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4 Aqueous Reactions Solutions to Exercises (b) MgSO4 1.3 10⁻² M MgSO4 = 1.3 10⁻² M 1.3 10⁻² M (c) C₆H₁₂O₆ (molecular solute); 0.0150 M C₆H₁₂O₆ = 0.0150 M C₆H₁₂O₆ (d) Plan. There is no reaction between NaCl and (NH₄)₂CO₃, so this is just a dilution problem, = Then account for ion stoichiometry. Solve. 45.0 mL + 65.0 mL = 110.0 mL total volume 0.272 M NaCl 45.0 mL = 0.111 M NaCl; 0.111 M Na+, 0.111 M 110.0 mL 0.0247 M (NH₄)₂ CO₃ X 65.0 mL = 0.0146 M (NH₄)₂CO₃ 110.0 mL 2x (0.0146 M) = 0.0292 M 0.0146 M Check. By adding the two solutions (with no common ions or chemical reaction), we have approximately doubled the solution volume, and reduced the concentration of each ion by approximately a factor of two. 4.72 (a) Plan. These two solutions have common ions. Find the ion concentration resulting from each solution, then add. Solve. total volume = 42.0 mL + 37.6 mL = 79.6 mL 0.170 M NaOH X 42.0 mL = 0.08970 = 0.0897 M NaOH; 79.6 mL 0.0897 M Na+, 0.0897 M OH⁻ 0.400 M NaOH 37.6 mL = 0.18894 = 0.189 M NaOH; 79.6 mL 0.189 M Na+, 0.189 M M Na+ = 0.08970 M + 0.18894 M = 0.27864 = 0.2786 M Na+ M = M Na+ = 0.2786 M (b) Plan. No common ions; just dilution. Solve. 44.0 mL + 25.0 mL = 69.0 mL 0.100 M Na₂SO₄ X 44.0 mL = 0.06377 = 0.0638 M 69.0 mL 2 (0.06377 M) = 0.1275 = 0.128 M 0.0638 M 92