Answer to Question #55379 in General Chemistry for hashim

Question #55379
*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?
Expert's answer
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.

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