Simple Calculations Question (1 Viewer)

[Nakashima]

Solutions of lead(II) nitrate (0.080 M, 60 mL) and potassium iodide (0.080 M, 40 mL) are mixed. What amount (in mol) of PbI₂ precipitates?

Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃

Number of moles of lead(II) nitrate: 0.08 x 0.06 = 4.8 x 10^-3
Number of moles of potassium iodide: 0.08 x 0.04 = 3.2 x 10^-3

Molar ratio = 1:2 and lead(II) nitrate is in excess.

. : 1.6 x 10^-3 moles of lead nitrate forms.

That part was right according to the answers.

The next bit:

What is the final concentration of K⁺ ions remaining in solution after the reaction?

Okay, is it just me or is the answer just 3.2 x 10^-3? Since the molar ratio's 1:1?

The answer given is 10 times as much.

Someone please confirm that I'm right. Or show me why I'm wrong.

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[Riviet]

Hi there, haven't seen you around forums for a while. I'm pretty sure it's you who I'm thinking of.

We know total volume of precipitate is 60ml + 40ml = 0.1 L

So c=n/v
=(3.2 x 10^-3)/0.1
=3.2 x 10^-2 mol/L

 

[Nakashima]

Oh shit, they asked for concentration!

My, my, how terribly embarassing.

Thanks Rivvy!