Thank you in advance Part A Calculate the number of mol of A^+ in 5.0 mL of 0.00

Question

Thank you in advance

Part A Calculate the number of mol of A^+ in 5.0 mL of 0.0040 M AgNO3 and the number of moles of CrO42^- in 5.0 ml of 0.0024 M K2CrO4. Use 2 sig fig. Separate your entries with a comma. Part B Write the solubility equilibrium net ionic equation for AgNO3 and K2CrO4. Include all phases. Part C If the solutions in part A are mixed. and the equilibrium is shifted towards completion which is the limiting reactant, Ag^+ orCrO4^2- ? Express your answer as chemical formula of the limiting reactant. Part D The Ksp for BaCrO4 is 1.2 x 10^-10. 10 mL of 1 x 10 ^-5 Ba(NO3)2 are mixed with 10 mL of 1 x 10 M K2CrO4? What is the ion product, Q, for BaCrO4? Use 2 sig fig. Part E Will BaCrO4 precipitate?

Explanation / Answer

Part A

No of mol of AgNo3 = 5/1000*0.004 = 2*10^-5 mol

1 mol AgnO3 = 1 mole Ag+

no of mol of Ag+ = 2*10^-5 mol

no of mol of K2CrO4 = 5/1000*0.0024 = 1.2*10^-5 mol

1 mol K2CrO4 = 1 mol CrO4^2-

no of mol of CrO4^2- = 2*1.2*10^-5 = 2.4*10^-5 mol

Part B

Molecular equation : AgNO3(aq) +K2CrO4(aq) ---> Ag2CrO4(S) + 2KNO3(aq)

inoic equation : Ag+(aq)+NO3-(aq) +2K+(aq)+CrO4^2-(aq) ---> Ag2CrO4(S) + 2K+(aq)+2NO3^-(aq)

net ionic equation : 2Ag+(aq) + CrO4^2-(aq) ----> Ag2CrO4(S)

Part C

2AgNO3(aq) +K2CrO4(aq) ---> Ag2CrO4(S) + 2KNO3(aq)

from equation

2 mol AgNo3 = 1 mol K2CrO4

no of mol of K2CrO4 = 5/1000*0.0024 = 1.2*10^-5 mol

No of mol of AgNo3 = 5/1000*0.004 = 2*10^-5 mol

limiting reagent is AgNo3 = Ag+

Part D

Ba(NO3)2(aq) + K2CrO4(aq) ----> BaCrO4(S) + 2 KNO3(aq)

Limiting reagent is Ba(NO3)2

concentration of BaCrO4 = 5*10^-6 M

Q = [Ba2+][CrO4^2-]

= (5*10^-6)*(5*10^-6) = 2.5*10^-11

Part E

No precipitate, because Q<Ksp

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