Which response has both answers correct? Will aprecipitate form when 250mL of .3

Question

Which response has both answers correct? Will aprecipitate form when 250mL of .33 MNa2CrO4 are added to 250mL of .12 MAgNO3? (Ksp(Ag2CrO4) =1.1x10-12). What is the concentration of the silver ionremaining in solution?
A) Yes, [Ag+] = 2.9x10-6 M B) Yes, [Ag+] = .060 M C) Yes, [Ag+] =1.3x10-4 M D) No, [Ag+] = .060 M E) No, [Ag+] = .105 M
Thanks for your help in advance! What is the concentration of the silver ionremaining in solution?
A) Yes, [Ag+] = 2.9x10-6 M B) Yes, [Ag+] = .060 M C) Yes, [Ag+] =1.3x10-4 M D) No, [Ag+] = .060 M E) No, [Ag+] = .105 M
Thanks for your help in advance!

Explanation / Answer

Na2CrO4 + 2AgNO3------> Ag2CrO4 +2NaNO3 No . of moles of Na2CrO4 , n = Molarity* Volume in L                                               = 0.33 M * 0.25 L                                               = 0.0825 moles No . of moles of AgNO3 , n' = 0.12 m * 0.25 L                                             =0.03 moles According to the Equation , 1 mole of Na2CrO4 reacts with 2moles of AgNO3 X moles of Na2CrO4 reactswith 0.03 moles of AgNO3 X = ( 0.03*1) / 2 = 0.015 So 0.0825 - 0.015= 0.0675 moles ofNa2CrO4 left unreacted 2 moles of AgNO3 produces 1 mole ofAg2CrO4 0.03 moles of AgNO3 produces 0.03/2 =0.015 moles of Ag2CrO4 No . of moles of Ag2CrO4 , n =Molarity * Volume of solution in L So Molarity of Ag2CrO4 formed is M= 0.015 moles / (0.25+0.25) L                                                           = 0.03 M So ionic productof  Ag2CrO4 is = [Ag2+ ]2 [ CrO4 2- ]                                                  = (2*0.03) 2 * 0.03                                                   =1.08*10^-4 Given solubility product , Ksp = 1.1x10-12 ===> Ionic product > Ksp So precipitate is formed Ag2CrO4 ---> 2Ag+ + CrO42- Concentration of Ag+ is 2 *0.03 = 0.06M X = ( 0.03*1) / 2 = 0.015 So 0.0825 - 0.015= 0.0675 moles ofNa2CrO4 left unreacted 2 moles of AgNO3 produces 1 mole ofAg2CrO4 0.03 moles of AgNO3 produces 0.03/2 =0.015 moles of Ag2CrO4 No . of moles of Ag2CrO4 , n =Molarity * Volume of solution in L So Molarity of Ag2CrO4 formed is M= 0.015 moles / (0.25+0.25) L                                                           = 0.03 M So ionic productof  Ag2CrO4 is = [Ag2+ ]2 [ CrO4 2- ]                                                  = (2*0.03) 2 * 0.03                                                   =1.08*10^-4 Given solubility product , Ksp = 1.1x10-12 ===> Ionic product > Ksp So precipitate is formed Ag2CrO4 ---> 2Ag+ + CrO42- Concentration of Ag+ is 2 *0.03 = 0.06M 0.03 moles of AgNO3 produces 0.03/2 =0.015 moles of Ag2CrO4 No . of moles of Ag2CrO4 , n =Molarity * Volume of solution in L So Molarity of Ag2CrO4 formed is M= 0.015 moles / (0.25+0.25) L                                                           = 0.03 M So ionic productof  Ag2CrO4 is = [Ag2+ ]2 [ CrO4 2- ]                                                  = (2*0.03) 2 * 0.03                                                   =1.08*10^-4 Given solubility product , Ksp = 1.1x10-12 ===> Ionic product > Ksp So precipitate is formed Ag2CrO4 ---> 2Ag+ + CrO42- Concentration of Ag+ is 2 *0.03 = 0.06M

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