Question

1. Will a precipitate form when 50.0 mg of sodium chloride is added to 250.0 mL of a 0.100 M solution of silver nitrate? If so, how many grams of precipitate form?

Clearly show all your calculations.

Answer 1

Here we know that percipitate will only for when Qsp > Ksp , so we have to calculate Qsp for Agcl because first lets see what will form when sodium chloride mixed with silver nitrate.

AgNo3 + Nacl -------> Agcl(s) + NaNo3

HERE Agcl------> Ag+ + cl-

which means Qsp = [Ag+][cl-]

Now mass given for Nacl = 50.0 mg= 50 x 10^-3 g

so No. of moles = 0.855 x 10^-3 moles

so [Nacl]= moles/ volume=0.855 x 10^-3/0.250 = 0.00342 M

therefore Na+=0.00342 M and Cl- =0.00342 M

now Agno3-----> Ag+ + No3-

[Ag+}=0.100 M

now Qsp = [Ag+][cl-]=0.100 x 0.00342=0.000342

ksp for Agcl= 1 × 10-10 .

so clearly qsp> ksp

therefore ppt will formed.

Now for mass we have

moles of agno3= 250 x 0.100/ 1000= 0.250

now SINCE moles of nacl< moles of agno3 therefore nacl will be limiting reagent

so 1 mole of nacl give 1 mol of agcl

therefore 0.855 x 10^-3 moles of Nacl will give 0.855 x 10^-3 moles of agcl

therefore mass = moles x molar mass = 0.855 x 10^-3 x143.32=122.53 x 10^-3=0.1225 gram ppt ANS

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