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
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.
1 answer