Will AgIO3 precipitate when 20 mL of 0.010M AgIO3 is mixed with 10 mL of 0.015M NaIO3? Ksp of AgIO3 is 3.1× 10-8
Will AgIO3 precipitate when 20 mL of 0.010 M AgNO3 is mixed with 10 mL of 0.015 M NaIO3?
Ksp of AgIO3 is 3.1×10−8
Solution:
AgNO3 → Ag+ + NO3−
So, the molarity of Ag+ is same as the molarity of AgNO3 (0.010 M)
NaIO3 → Na+ + IO3−
So, the molarity of IO3− is same as the molarity of NaIO3 (0.015 M)
Moles of Ag+ = (0.010 M) × (20 mL) × (1L / 1000 mL) = 0.0002 mol
Moles of IO3− = (0.015 M) × (10 mL) × (1L / 1000 mL) = 0.00015 mol
Total volume = 20 mL + 10 mL = 30 mL = 0.03 L
Therefore,
[Ag+] = (0.0002 mol) / (0.03 L) = 0.00667 M
[IO3−] = (0.00015 mol) / (0.03 L) = 0.0050 M
AgIO3 → Ag+ + IO3−
Qsp = [Ag+] × [IO3−] = (0.00667) × (0.0050) = 3.33×10−5
Since Qsp > Ksp
3.33×10−5 > 3.1×10−8
Therefore,
AgIO3 will precipitate
Answer: AgIO3 will precipitate
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