Calculations Involving Solutions UNIT VIII PPT #2 Calculations Involving Solutions

IX.6 CALCULATING THE CONCENTRATIONS OF IONS IN SOLUTIONS REVIEW: M = [ ] = molarity = C = n = mol = molar concentration V L

IX.6 CALCULATING THE CONCENTRATIONS OF IONS IN SOLUTIONS 1.What is the concentration of Cl ions in 0.350M solution of HCl? HCl → H+ + Cl- [Cl-] = 0.350 mol HCl x 1 mol Cl- = 0.350 M Cl- L 1 mol HCL 2. What is the concentration of H= ions in a 1.40 M solution of H2SO4? H2SO4 → 2H+ + SO42- [H+] = 1.40M H2SO4 x 2 mol H+ = 2.80 M H+ L 1 mol H2SO4

IX.6 CALCULATING THE CONCENTRATIONS OF IONS IN SOLUTIONS Example #1: a. What is the molar concentration of the sodium ions in 0.25M NaOH(aq)? NaOH → Na+ + OH- [Na+] = 0.25 mol NaOH x 1 mol Na+ = 0.25 M 1L 1 mol NaOH b. What is the molar concentration of the chloride ions in 0.45 M AlCl3(aq)? AlCl3 → Al3+ + 3Cl- [Cl-] = 0.45 mol AlCl3 x 3 mol Cl- = 1.35 M 1L 1 mol AlCl3

IX.6 CALCULATING THE CONCENTRATIONS OF IONS IN SOLUTIONS Example #2 What is the chloride ion concentration when 20.0 g of BaCl2(s) is dissolved and diluted to 800.mL? BaCl2 → Ba2+ + 2Cl- 1. # mol = 20. 0 g x 1 mol = 0.096 mol 208.3 g 2. C=n/V = 0.096 mol/0.800L = 0.12 M 3. [Cl-] = 0.12 mol x 2 mol Cl- = 0.24 M 1L 1 mol BaCl2

IX.6 CALCULATING THE CONCENTRATIONS OF IONS IN SOLUTIONS REVIEW: C1V1 = C2V2 Cdil x Vdil = Cconc x Vconc

IX.6 CALCULATING THE CONCENTRATIONS OF IONS IN SOLUTIONS Example #3 A solution is made by mixing 100.0 mL of 0.250 M BaSO4 and 80.0 mL of 0.500 M MgF2. What is the concentration of each ionic species in the final solution? Steps: Balanced dissociation reactions Dilutions to find final Molarity of both Set up concentrations with dissociation equation ID species (and multiply)

IX.6 CALCULATING THE CONCENTRATIONS OF IONS IN SOLUTIONS Example #4 A solution is made by mixing 50.0 mL of 0.300 M BaCl2 and 100.0 mL of 0.500 M NaCl. What is the concentration of each ionic species in the final solution? Hebden p. 212 # 30-38