I can't figure out this stoichiometry question!

10.0 mL of 1.15M copper(II) nitrate reacted with 20.0 mL of .85M ammonium hydroxide. The solution is filtered. Determine the molarity of the filtrate.

I figured out that the net ionic equation is Cu + 2OH -> Cu(OH)2, so the 2NH4NO3 dissolves, and the Cu(OH)2 probably doesn't pass through the filter, but I don't know where to go from there. Thanks in advance!

  • $\begingroup$ The question is poorly worded since it doesn't specify for which chemical species the molarity is desired. Copper(II) would seem to be the species of interest. $\endgroup$ – MaxW Mar 15 '16 at 22:37

$$\ce{Cu(NO3)2}+2\ce{NH4OH} =>\ce{Cu(OH)2}(s)+2\ce{NH4NO3} $$ Essentially, we're trying to find the Molarity of $\ce{NH4NO3}$ since the copper filters out(as you stated)

First solve for limiting reagent. $$10mL\ce{Cu(NO3)2}*\frac{1L}{1000mL}*\frac{1.15mol}{1L}=0.0155mol\ce{Cu(NO3)2}$$ $$20mL\ce{NH4OH}*\frac{1L}{1000mL}*\frac{0.85mol}{1L}=0.0170mol\ce{NH4OH}*\frac{1}{2}=0.0085$$

  • Why do we divide mols of $\ce{NH4OH}$ by 2? Remember that for every molecule of $\ce{Cu(NO3)2}$, there must be 2 molecules of $\ce{NH4OH}$. $$\frac{2\ce{NH4OH}}{\ce{Cu(NO3)2}}$$ for this reason we need to divide the amount of $\ce{NH4OH}$ by 2 to compensate for stoicheometry. So actually, $\ce{NH4OH}$ our limiting reagent. Since you didn't give my any solubility constants, I'm going to assume 100% of the Copper became $\ce{Cu(OH)2}(s)$ and all of the ammonium became $\ce{NH4NO3}$. In the real world, perciptates never form at 100%.

0.0085 mols of $\ce{Cu(OH)2}$ is produced. You want $\ce{NH4NO3}$ Can you use stoichometric ratios to proceed from there? Molarity - (mol/L). Make sure you add up the two volumes to find the Total volume.

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  • $\begingroup$ You assumption that all the copper was converted to copper hydroxide is wrong. $\endgroup$ – MaxW Mar 15 '16 at 23:38
  • $\begingroup$ Yeah I know. but usually in HS chemistry when students are learning stoichometry, they generally ignore Solubility constants and whatnot. I'll edit my post a little. $\endgroup$ – Jess L Mar 15 '16 at 23:40
  • $\begingroup$ I still can't figure it out. I found that .83 grams of Cu(OH)2 was produced using stoichiometry, but I don't think this would help. I don't know what to do with the Total volume of 30 mL. I also figured out that .003 moles of excess reactant remains after the reaction, so that might affect my answer. The answer key says the answer is .10M $\endgroup$ – Alex S Mar 16 '16 at 0:10
  • $\begingroup$ Whoops. That's embarrassing- looks like I didn't help much. In that case use the numbers you got. 0.003mol/30mL= .003mol/.03L . =0.10 mol/L = molarity of excess reactant in solution. $\endgroup$ – Jess L Mar 16 '16 at 0:19
  • $\begingroup$ Thank you! I think I was confused about what molarity meant (since we learned about it a while ago), but now it makes sense. $\endgroup$ – Alex S Mar 16 '16 at 0:31

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