Answer to Question #55379 in General Chemistry for hashim
*How many moles of BaCl2 are formed in the neutralization of 196.5 mL of 0.095 M Ba(OH)2 with aqueous HCl?
*Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pb2+(aq) + 2I-(aq) → PbI2(s)
Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 1.180 M HI(aq) must be added to a solution containing 0.200 mol of Pb(NO3)2(aq) to completely precipitate the lead?
*What is the molarity of a NaOH solution if 15.5 mL of a 0.220 M H2SO4 solution is required to neutralize a 25.0 mL sample of the NaOH solution?
Solution: C=Ʋ/V, [mol/L] *196.5 mL = 0.1965 L 0.1965 L × 0.095 mol/L = 0.0186675 mol = 0.0187 mol *0.200 mol / 1.180 mol/L = 0.1695 L = 169.5 mL * 0.0155 L × 0.220 mol/L = 0.00341 mol 0.00341 mol / 0.025 L = 0.1364 M Answer: *0.0187 moles of BaCl2; *169.5 milliliters of 1.180 M HI; *0.1364 M NaOH.