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

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The molar concentration of ions in a 0.16M solution of Pb(NO3)2 is determined by its dissociation into Pb^2+ and 2 NO3^-. For every mole of Pb(NO3)2, one mole of Pb^2+ and two moles of NO3^- are produced, resulting in a Pb^2+ concentration of 0.16M and a NO3^- concentration of 0.32M. The calculations presented in the discussion incorrectly focus on mass rather than the stoichiometric relationships of the ions. The correct approach is to directly apply the molarity of the compound to find the ion concentrations. Thus, the final concentrations are 0.16M for Pb^2+ and 0.32M for NO3^-.
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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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