A 50 mL solution of 0.2 M Pb(NO3)2 is titrated with a 5 x 10^-2 M KI solution. T

Question

A 50 mL solution of 0.2 M Pb(NO3)2 is titrated with a 5 x 10^-2 M KI solution. The first appearance of a yellow color that lasts for more than 30 seconds occurs after 14.50 mL KI have been added. What is the value of the ion product of the titration?

Explanation / Answer

follow this ) M = m/MM x V, where m = mass, MM = molar mass, V = volume in liters MM = m/MxV ---> 240-g/0.50-mol/L x 4-L = 120 g/mol Ans = 120 is molecular mass of unknown 2. Since we want 1/2-L of 2M solution we need 1 mole of Na+ in 1/2 liter of solution. Each formula unit of the substance, Na2SO4, contains 2 sodium atoms so we need 0.5 moles of Na2SO4 in 500-mL M = mol / V ---> 2.0 mol/L = ?-mol Na2SO4 x (2 mol Na+/ 1 mol Na2SO4)/ .500-L = .5 mol needed Ans= 0.5 moles of Na2SO4 needed 3. 0.2 mole I- + 0.4 mole I= = 0.6 mole I-. To complex all of 0.6 moles into PbI2 would require 0.3 moles of Pb(C2H3O2)2 Ans = 0.3 moles Pb(C2H3O2)2 needed 4. Pb(NO3)2 + MgCl2 ---> PbCl2 + Mg(NO3)2. a.) moles of Pb+2 = 0.20-M x .300-L = 0.06 mol Pb+2 b.) moles reactant MgCl2 = 0.2M x .200-L = 0.04 mol c.) this leaves 0.02 moles of Pb+2 unreacted d. final concentration of Pb+2 = 0.02 mol / .5-L = 0.04 M Pb+2 Ans = final Pb+2 concentration = 0.04 M

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