3. A pipette is used to add 46 mL of 0.1 M Pb(NO3)2 to a 294.0...

3. A pipette is used to add 46 mL of 0.1 M Pb(NO3)2 to a 294.0 mL solution of 0.2 M NaF. Determine the equilibrium concentration of Pb2+(aq) in the final solution. Ksp(PbF2)=3.6E-8

Expert Solution

queen_honey_blossom answered 2 years ago

If we mix 19.0 mL of a 1.30 ✕ 10−3 M solution of Pb(NO3)2 with 18.6...

If we mix 19.0 mL of a 1.30 ✕ 10−3 M solution of Pb(NO3)2 with 18.6 mL of a 2.60 ✕ 10−3 M solution of NaBr, will a precipitate form? At this point we would identify the potential precipitate by switching the anions and using the solubility rules to decide if one of the potential products is potentially insoluble. Here the products would be PbBr2 and NaNO3. Of these, PbBr2 is considered "insoluble". The Ksp value for PbBr2 is 6.60 ...

A solution is prepared by mixing 50 mL of 1 M Pb(NO3)2 and 75 mL of...

A solution is prepared by mixing 50 mL of 1 M Pb(NO3)2 and 75 mL of .5 M NaF. Calculate the concentration of F- ions present at equilibrium. (Hint: First, write and balance the double displacement reaction taking place between Pb(NO3)2 and NaF to form PbF2(s). Perform the necessary stoichiometry (including finding which reactant is limiting) to calculate how much of each reactant remains after the reaction goes to completion. Remember that stoichiometry has to be done in moles. Once...

A solution is prepared by mixing 100 mL of .01 M Pb(NO3)2 and 100 mL of...

A solution is prepared by mixing 100 mL of .01 M Pb(NO3)2 and 100 mL of .001 M NaF. Will PbF2(s) precipitate in this reaction? Calculate the concentration of F- ions present at equilibrium. PbF2(s) <--> Pb2+(aq) + 2F-(aq) Kc= 3.7E-8 Please show all work! Thank you!

If 2.00 mL of 0.100 M Pb(NO3)2 are added to 1.00 L of 0.100 M KCl...

If 2.00 mL of 0.100 M Pb(NO3)2 are added to 1.00 L of 0.100 M KCl will a precipitate form? What will be the precipitate?

A student mixes 31.0 mL of 3.24 M Pb(NO3)2(aq) with 20.0 mL of 0.00217 M NaI(aq)....

A student mixes 31.0 mL of 3.24 M Pb(NO3)2(aq) with 20.0 mL of 0.00217 M NaI(aq). How many moles of PbI2(s) precipitate from the resulting solution? What are the values of [Pb2 ], [I–], [NO3–], and [Na ] after the solution has reached equilibrium at 25 °C?

1. a.) A student mixes 47.0 mL of 3.30 M Pb(NO3)2(aq) with 20.0 mL of 0.00183...

1. a.) A student mixes 47.0 mL of 3.30 M Pb(NO3)2(aq) with 20.0 mL of 0.00183 M Na2C2O4(aq). How many moles of PbC2O4(s) precipitate from the resulting solution? What are the values of [Pb2 ], [C2O42–], [NO3–], and [Na ] after the solution has reached equilibrium at 25 °C? Ksp of PbC2O4 (s) is 8.5e^-9 M^2 b.) What are the values of [Pb^2+], [C2O4^2-], [NO3^-], and [Na^+] after the solution has reached equilibrium?

If we mix 40.0 mL of 3.00 M Pb(NO3)2 (aq) with 20.0 mL of 2.00 x...

If we mix 40.0 mL of 3.00 M Pb(NO3)2 (aq) with 20.0 mL of 2.00 x 10-3 M NaI (aq), does PbI2 (s) precipitate from solution? I yes, calculate how many moles of PbI2 (s) precipitate and the values of [Pb2+], [I-], [NO3-], and [Na+] at 25 C at equilibrium?

100.0 mL of 0.500 M lead(II) nitrate (Pb(NO3)2, 331 g/mol) and 100.0 mL of 0.500 M...

100.0 mL of 0.500 M lead(II) nitrate (Pb(NO3)2, 331 g/mol) and 100.0 mL of 0.500 M sodium sulfate (Na2SO4, 142 g/mol) are mixed together. (I) Write the overall balanced equation. Include the states of each species. (II) Write the overall (or total) ionic equation for the reaction. Identify any spectator ions. Include states of each species. (III) Write the net ionic equation. Include states of each species. (IV) Determine the limiting reactant and the theoretical yield of the non-aqueous product....

Reaction is: Pb(NO3)2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO3(aq) 2.0 ml of 0.250M Pb(NO3)2 and 7.0...

Reaction is: Pb(NO3)2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO3(aq) 2.0 ml of 0.250M Pb(NO3)2 and 7.0 ml of 0.250M KI A) In your experiment, one of the reagents was limiting and the other was in excess. Therefore after the lead(II)iodide precipitated from the solution, there were spectator ions left in the solution, and also the unreacted excess reagent. Assuming complete precipitation of lead (II) iodide, calculate the concentrations of each type of ion (Pb2+, K+, NO3-, I-) present in the...

A solution is 0.010 M in each of Pb(NO3)2, Mn(NO3)2, and Zn(NO3)2. Solid NaOH is added...

A solution is 0.010 M in each of Pb(NO3)2, Mn(NO3)2, and Zn(NO3)2. Solid NaOH is added until the pH of the solution is 8.50. Which of the following is true? Pb(OH)2, Ksp= 1.4 x10^-20 Mn(OH)2, Ksp= 2.0 x 10^-13 Zn(OH)2, Ksp= 2.1 x10^-16 A) No precipitate will form. B) Only Pb(OH)2 will precipitate. C) Only Mn(OH)2 wil precipitate. D) Only Zn(OH)2 and Pb(OH)2 will preipitate. E) All three hydroxide will preipitate.

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