Question

770. mL of 0.00639 M NaI (aq) is combined with 285. mL of 0.00146 M Pb(NO3)2 (aq). Determine if a precipitate will form given that the Ksp of Pbl2 is 1.40x10-8. Precipitation will occur because Qsp < Ksp Precipitation will not occur because Qsp > Ksp Precipitation will occur because Qsp = Ksp Precipitation will occur because Qsp > Ksp Precipitation will not occur because Qsp < Ksp

Answer 1

Answer:

Precipitation will not occur because Qsp < Ksp

Explanation:

Step 1: Data given

Volume of 0.00639 M NaI = 770 mL

Volume of 0.00146 M Pb(NO3)2 = 285 mL

Ksp of PbI2 = 1.40 *10^-8

Step 2: The balanced equation

Pb(NO3)2 + 2NaI → PbI2 + 2NaNO3

Step 3: Calculate new concentrations

[NaI] = 0.00639 * (770/1055)

[NaI] = 0.004664 M

[Pb(NO3)2] = 0.00146 * (285/1055)

[Pb(NO3)2] = 0.0003944 M

Step 4: Calculate Q

Q(pbI2) = [Pb2+]*[I-]² = [Pb(NO3)2] * [NaI]²

Q = 0.0003944 * 0.004664²

Q =  0,0000000085793421824 = 8.58 *10^-9

Q < Ksp

This means a precipitate will not be formed

Precipitation will not occur because Qsp < Ksp