Question

How many micrograms of sodium chloride (to 2 decimal places) can be added to 949 mL of 0.00150 M silver nitrate before a precipitate forms? (K_sp= 1.80 x 10^-10)

Answer 1

6.66micro grams

Explanation

Concentration of AgNO3 = 0.0015M

So, Concentration of Ag+ = 0.0015M

equillibrium of solubility of AgCl

AgCl(s) <-------> Ag+(aq) + Cl-(aq)

Ksp = [Ag+][Cl-]= 1.80×10^-10

[Ag+][Cl-] = 1.80×10^-10

substituting value for [Ag+]

0.0015M [ Cl-] = 1.80×10^-10M^2

[ Cl-] = 1.80×10^-10M^2/0.0015M

= 1.200×10^-7M

So, when [Cl-] going to abovev1.200×10^-7M , precipitate will start

[ Cl- ] = [NaCl]

Volume of solution = 949ml

No of mole of NaCl = (1.200×10^-7mole/1000ml)×949=1.1388×10^-7mole

Molar mass of NaCl = 58.44g/ml

mass of NaCl =1.1388×10^-7mole × 58.44g/mole =66.55×10^-7g = 6.66micro grams