Pb(NO3)2 0.16M: Molar Concentration of Ions

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SUMMARY

The molar concentration of ions in a 0.16M solution of Pb(NO3)2 is determined by the dissociation of the compound into its constituent ions. Each formula unit of Pb(NO3)2 produces one Pb2+ ion and two NO3- ions. Therefore, the concentration of Pb2+ is 0.16M, while the concentration of NO3- is 0.32M. The calculations presented in the discussion incorrectly attempted to derive these concentrations using molar masses and ratios, leading to confusion.

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Homework Statement



What molar concentration of each ion appear in 0.16M Pb(NO3)2?

*Note* The title incorrectly says "moles".

Homework Equations



M = mol/L

The Attempt at a Solution



My teacher tells me this is wrong, but I can't see my fault:

Molar Mass of Pb+2 = 207.20g
Molar Mass of NO3- = 124.02g

\frac{x}{1L} = 0.16M

x = 0.16mol

0.16mol = \frac{y*1mol}{331.22g}

y = 53gPb: 53g * \frac{207.20g}{331.22g} = 33g

\frac{33g}{331.22g} = 0.10molNO_{3}: 53g *\frac{124.02}{331.22g} = 20g

\frac{20g}{331.22g} = 0.06molPb = \frac{0.10mol}{1L} = 0.10M

NO_{3} = \frac{0.06mol}{1L} = 0.06M

Any ideas where I messed up?
Or if I did?
 
Last edited:
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I would also like to say that in my first attempt to solve this, I simply multiplied 0.16M by the fractions of mass for lead and nitrate and got the same answers.

That was the "incorrect" version. This is my third attempt, with the same results.
 

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