This is the oxidation cr 2 o 7 2 6e 14h 2cr 3to have

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(this is the oxidation) Cr 2 O 7 2– + 6e + 14H + 2Cr 3+ (to have +6 = +6, charge) Cr 2 O 7 2– + 6e + 14H + 2Cr 3+ + 7H 2 O (to balance H & O) Cr 2 O 7 2– + 6e + 14H + 2Cr 3+ + 7H 2 O to equal # of 6 × (Fe 2+ Fe 3+ + e ) electrons, 6 Cr 2 O 7 2– + 14H + + 6Fe 2+ 2Cr 3+ + 7H 2 O + 6 Fe 3+
Balancing redox equations ON: +7 +4 in basic media (OH ) +4 +6 MnO 4 (aq) + SO 3 2 (aq) MnO 2 (s) + SO 4 2 (aq) MnO 4 + 3e MnO 2 (this the reduction) SO 3 2 SO 4 2 + 2e (this is the oxidation) MnO 4 + 3e MnO 2 + 4OH (to equal charges, 4) SO 3 2 + 2OH SO 4 2 + 2e (to equal charges, 4) 2 × (MnO 4 + 3e +2H 2 O MnO 2 + 4OH ) to equal H, O, 3 × (SO 3 2 + 2OH SO 4 2 + 2e + H 2 O) and electrons 2MnO 4 + H 2 O + 3SO 3 2 2MnO 2 + 2OH + 3SO 4 2
Measuring Concentrations of Compounds in Solutions Concentration Terms
Parts Per Hundred (percent, %) weight/weight, %(w/w) ( most common ) mass solute (g) ———————— x 100 mass solution (g) volume/volume, %(v/v) V solute (mL) liquid solute in ———————— x 100 liquid solvent V solution (mL)
Parts Per Hundred (percent, %) weight/volume, %(w/v) mass solute (g) ———————— x 100 V solution (mL)
Learning Check A solution is prepared by mixing 15.0 g of Na 2 CO 3 and 235 g of H 2 O. The final V of solution is 242 mL. Calculate the %w/w and %w/v concentration of the solution. g solution = 15.0 g Na 2 CO 3 + 235 g H 2 O = 250. g 15.0 g solute %w/w = ——————— x 100 = 6.00% Na 2 CO 3 250. g solution 15.0 g solute %w/v = ——————— x 100 = 6.20% Na 2 CO 3 242 mL solution
Molarity, M The Molarity, M , usually known as the molar concentration of a solute in a solution, is the number of moles of solute per liter (1000 mL) of solution / or mmoles per mL of solution. To calculate it we need moles of solute and V(in liters) of solution (or mmol and mL) and to divide mol solute mmol solute M = —————— = ——————— V(L) solution V(mL) solution
Calculation of Molarity, M What is the molarity of 500. mL NaOH solution if it contains 6.00 g NaOH? 500. mL ÷1000 = 0.500 L (volume in liters) FW (NaOH) = 40.0 g/mol (from periodic table) How many moles of NaOH?
Example: Describe the preparation of 300.0 mL of 0.4281 Msilver nitrate solution. 300.0 mL ÷1000 = 0.3000 L AgNO3FW = 169.97 g/mol 0.4281 mol 0.3000 L x ————— = 0.1284 mol AgNO31 L soln 169.97 g

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