4) Using your initial mass of copper (0.38), calculate the minimum volume of concentrated 12M nitric acid needed to completely react with your copper to form copper (II) nitrate
5) Using your answer from question 4 (2x10^-3 L), calculate the minimum volume of 6M sodium hydroxide solution required to completely precipitate all of the copper as copper (II) hydroxide.
You need help with 5.
mols Cu you have from 4 (0.38/63.5) = ?
Cu(NO3)2 + 2NaOH ==> 2NaNO3 + Cu(OH)2
You know mols Cu and that's the same as mols Cu(NO3)2.
mols NaOH = twice that
Theh L NaOH = mols NaOH/M NaOH and solve for L
To calculate the minimum volume of concentrated 12M nitric acid needed to completely react with the copper, you can use the concept of mole-to-mole ratios.
First, you need to determine the number of moles of copper you have. The molar mass of copper is 63.55 g/mol, so the number of moles can be calculated using the formula:
moles of copper = mass of copper / molar mass of copper
For example, let's say your initial mass of copper is 0.38 g:
moles of copper = 0.38 g / 63.55 g/mol ≈ 0.006 mol
Since the balanced chemical equation for the reaction between copper and nitric acid is:
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
we can see that the mole ratio between copper and Cu(NO₃)₂ is 1:1.
Thus, the number of moles of Cu(NO₃)₂ formed will be the same as the number of moles of copper.
Now let's calculate the volume of concentrated nitric acid needed.
moles of Cu(NO₃)₂ = moles of copper = 0.006 mol
Using the given concentration of the nitric acid (12M), we can set up the following equation:
moles of Cu(NO₃)₂ = concentration of nitric acid × volume of nitric acid
0.006 mol = 12 mol/L × volume of nitric acid
Solving for the volume of nitric acid:
volume of nitric acid = 0.006 mol / 12 mol/L ≈ 5 x 10⁻⁴ L or 5 microliters (µL)
Therefore, the minimum volume of concentrated 12M nitric acid needed to completely react with 0.38 g of copper to form copper (II) nitrate is approximately 5 microliters (µL).
Now let's move on to question 5.
To calculate the minimum volume of 6M sodium hydroxide solution required to completely precipitate all of the copper as copper (II) hydroxide, we can use the same mole-to-mole ratio concept and the volume of copper (II) nitrate from question 4.
The balanced chemical equation for the precipitation reaction is:
Cu(NO₃)₂ + 2NaOH → Cu(OH)₂ + 2NaNO₃
From the balanced equation, we can see that the mole ratio between Cu(NO₃)₂ and Cu(OH)₂ is 1:1.
Since we calculated the moles of Cu(NO₃)₂ in question 4 as 0.006 mol, we need an equal amount of moles of Cu(OH)₂ to completely precipitate the copper.
Using the given concentration of sodium hydroxide solution (6M), we can set up the following equation:
moles of Cu(OH)₂ = concentration of sodium hydroxide × volume of sodium hydroxide
0.006 mol = 6 mol/L × volume of sodium hydroxide
Solving for the volume of sodium hydroxide:
volume of sodium hydroxide = 0.006 mol / 6 mol/L = 0.001 L or 1 milliliter (mL)
Therefore, the minimum volume of 6M sodium hydroxide solution required to completely precipitate all of the copper as copper (II) hydroxide is 1 milliliter (mL).