ICE Challenge

1. 25. 0 mL of 0. 100 M NaOH, 10. 0 mL 0. 200 M KOH, and 20. 0 mL of 0 1. 25.0 mL of 0.100 M NaOH, 10.0 mL 0.200 M KOH, and 20.0 mL of 0.100 M H2SO4 are poured into the same beaker. What is the resulting concentration of the excess acid or base? 0.0250 L x 0.100 mole NaOH = 0.00250 mol L 0.0100 L x 0.200 mole KOH = 0.00200 mol = 0.00450 mol Total Base 0.0200 L x 0.100 mole H2SO4 = 0.00200 mol Total Acid 2XOH + H2SO4 → X2SO4 + 2HOH I 0.00450 mol 0.00200 mol C 0.00400 mol 0.00200 mol E 0.00050 mol 0.00000 Total Volume = 25.0 mL + 10.0 mL + 20.0 mL = 55.0 mL Molarity Bases = 0.00050 mol = 0.0091 M 0.0550 L

2. 250.0 mL of 0.100 M HCl and 100.0 mL 0.200 M HNO3, react with excess CaCO3. What is the resulting volume in mLof CO2 produced at STP. 2HCl + CaCO3 → CO2 + CaCl2 + H2O 0.250 L HCl x 0.100 mole x 1 mole CO2 x 22.4 L x 1000 mL = 280 mL 1 L 2 mole HCl 1 mole 1L 2HNO3 + CaCO3 → CO2 + Ca(NO3)2 + H2O 0.100 L HNO3 x 0.200 mole x 1 mole CO2 x 22.4 L x 1000 mL = 224 mL 1 L 2 mole HNO3 1 mole 1L Total CO2 = 504 mL