CHEMISTRY

posted by .

Wondering If I did this correctly so far
50.0 ml of a 0.0500 M solution of lead (II) nitrate is mixed with 40.0 ml of a 0.200 M solution of sodium iodate at 25°C. Calculate the Pb2+ and IO3- concentrations when the mixture comes to equilibrium. At this temperature the Ksp for lead (II) iodate is 2.6 x 10-13. (10 marks)

The balanced equation
Pb(NO3)2 + 2NaIO3 ==> Pb(IO3)2 + 2NaNO3
Pb(NO3)2 = (0.05L)(0.05M) = 2.5 x 10-3 mols.
NaIO3 = (0.04L)(0.2M) = 8.0 x 10-3 mols.
NaIO3 is in excess.
Now if this is correct what do I do next?

So far so good.
NaIO3 is is excess; therefore, determine mols Pb(IO3)2 pptd and mols NaIO3 remaining.(All of the Pb(NO3)2 will be consumed.) Then it becomes a common ion problem with Ksp.
You will have solid Pb(IO3)2 in a saturated solution of Pb(IO3)2 that also contains xx mols NaIO3. Calculate Pb^+2.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chem

    I'm stuck on this one aswell... 50.0 ml of a 0.0500 M solution of lead (II) nitrate is mixed with 40.0 ml of a 0.200 M solution of sodium iodate at 25°C. Calculate the Pb2+ and IO3- concentrations when the mixture comes to equilibrium. …
  2. chemistry

    OKOK I know that I have posted this question but I just don't understand the last part. 50.0 ml of a 0.0500 M solution of lead (II) nitrate is mixed with 40.0 ml of a 0.200 M solution of sodium iodate at 25°C. Calculate the Pb2+ and …
  3. chemistry check 2

    I know you've read this question too many times already but I just want to make sure I did the last part (Ksp) correctly. 50.0 ml of a 0.0500 M solution of lead (II) nitrate is mixed with 40.0 ml of a 0.200 M solution of sodium iodate …
  4. chemistry

    a student synthesized 6.895g of barium iodate monohydrate, Ba(IO3)2.H2O, by adding 30.00 mL of 5.912*10^-1M barium nitrate, Ba(NO3), to 50.00 mL of 9,004*10^-1 M sodium iodate, NaIO3 (1) write the chemical equation for the reaction …
  5. Chemistry

    A solution containing Pb2+ and a solution containing IO3- are poured together and quickly mixed. After mixing, the solution contains 0.0030 M Pb2+ and 0.040 M IO3-. Immediately Pb(IO3)2(s), which has Ksp=2.510-13, begins to precipitate. …
  6. Chemistry(Please check)

    . Ksp for Fe(IO3)3 is 10-14. Two solutions, one being iron(III) nitrate and the other being sodium iodate, were mixed. At the instant of mixing, [Fe3+] = 10-4M and [IO3-] = 10-5M. What happens?
  7. Chemistry

    When reviewing the procedure the student found that 4.912 x 10-1 M barium nitrate solution had been used instead of that in the following procedure: A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by adding …
  8. Chemistry

    When reviewing the procedure the student found that 4.912 x 10-1 M barium nitrate solution had been used instead of that in the following procedure: A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by adding …
  9. Chem

    When 20.0mL of 1.40M solution of calcium nitrate is mixed with 70.0mL of 0.234M potassium iodate, a precipitate Ca (IO3)2 is formed. Calculate the concentrations of both NO3- and IO3- before mixing and after mixing.
  10. Chemistry

    When 20.0mL of 1.40M solution of calcium nitrate is mixed with 70.0mL of 0.234M potassium iodate, a precipitate Ca (IO3)2 is formed. Calculate the concentrations of both NO3- and IO3- before mixing and after mixing

More Similar Questions