# What is the amount of nitric acid required to oxidize a 0.1025*g mass of copper metal to Cu(NO_3)_2 if its VOLUME is 200*mL?

Apr 7, 2016

${\text{0.008065 mol L}}^{- 1}$

#### Explanation:

Before doing anything else, you need to write a balanced chemical equation for this redox reaction.

Nitric acid, ${\text{HNO}}_{3}$, is a very powerful oxidizing agent that will oxidize copper metal to copper(II) cations ,${\text{Cu}}^{2 +}$, while being reduced to nitrogen dioxide, ${\text{NO}}_{2}$, in the process.

Mind you, concentrated nitric acid will be reduced to nitrogen dioxide; by comparison, dilute nitric acid will be reduced to nitric oxide, $\text{NO}$.

So, the balanced chemical equation looks like this

${\text{Cu"_ ((s)) + 4"HNO"_ (3(aq)) -> "Cu"("NO"_ 3)_ (color(red)(2)(aq)) + 2"NO"_ (2(g)) uarr + 2"H"_ 2"O}}_{\left(l\right)}$

Copper(II) nitrate is soluble in aqueous solution, which means that it exist as cations and ions in solution. Likewise, nitric acid is a strong acid that dissociates completely in aqueous solution, so you can rewrite the equation as

${\text{Cu"_ ((s)) + 4"H"_ ((aq))^(+) + 4"NO"_ (3(aq))^(-) -> "Cu"_ ((aq))^(2+) + color(red)(2)"NO"_ (3(aq))^(-) + 2"NO"_ (2(g)) uarr + 2"H"_ 2"O}}_{\left(l\right)}$

This will give you the net ionic equation for this reaction

${\text{Cu"_ ((s)) + 4"H"_ ((aq))^(+) + 2"NO"_ (3(aq))^(-) -> "Cu"_ ((aq))^(2+) + 2"NO"_ (2(g)) uarr + 2"H"_ 2"O}}_{\left(l\right)}$

Now, because you're dealing with concentrated nitric acid, it's safe to assume that the acid will be in excess and that all the moles of copper metal will take part in the reaction.

Use copper metal's molar mass to determine how many moles you have in that $\text{0.1025-g}$ sample

0.1025 color(red)(cancel(color(black)("g"))) * "1 mole Cu"/(63.546color(red)(cancel(color(black)("g")))) = "0.001613 moles Cu"

Since you have a $1 : 1$ mole ratio between copper metal and copper(II) cations, you can say that the reaction will produce $0.001613$ moles of copper(II) cations.

The total volume of the solution is said to be equal to $\text{200.0 mL}$. This means that the molarity of the copper(II)cations, which tells you how many moles of copper(II) cations you'd get per liter of this solution, will be equal to

["Cu"^(2+)] = "0.001613 moles"/(200.0 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"0.008065 mol L"^(-1)color(white)(a/a)|)))

I'll leave the answer rounded to four sig figs.