When 80.0 mL of a 0.110M Lead(II) Nitrate Solution is mixed with 118.5mL of a 0.

Question

When 80.0 mL of a 0.110M Lead(II) Nitrate Solution is mixed with 118.5mL of a 0.185M Potassium Iodide solution, a yellow-orange precipitate of of Lead(II) Iodide is formed

a) what mass in grams of lead(II) Iodide is formed, assuming the reaction goes to completion ?

b) What is the molarity of Pb2+ in the resulting Solution ?

c) what is the molarity of K+ in the Resulting solution ?

d) What is the molarity of NO3- in the resulting solution ?

e) What is the molarity of I- in the resulting solution ?

please if you wont explain , give me the right answers , these questions drove me crazy !

Explanation / Answer

80.0 mL of a 0.110M Lead(II) Nitrate = 8.80 x10^-3 mol of Pb2+ and 17.6 x10^-3 mol of NO3-

118.5mL of a 0.185M Potassium Iodide = 21.9x10^-3 mol each of K+ and I-

a)

Thus mol of lead(II) Iodide = 8.80 x10^-3 mol

molar mass of lead(II) Iodide = 461 g/mol

thus grams of lead(II) Iodide = 8.80 x10^-3 mol * 461 g/mol = 4.06 g

b)

molarity of Pb2+ = 0 M (all Pb 2+ is consumed)

c)

molarity of K+ = (21.9x10^-3 ) / (0.080+0.1185) = 0.110 M

d)

molarity of NO3- in the resulting solution =17.6 x10^-3 / (0.080+0.1185) = 0.0887 M

e)

molarity of I- = (21.9x10^-3 - 2*8.80 x10^-3) / (0.080+0.1185) = 0.0217 M

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