[Inojim]

First off, I'm not a chemist at all. But I want to duplicate an electroplating experiment and needed some basic help. I put my question on a couple of forums and got two answers.

My question had two parts.  First, how to make a 0.01mol/L solution of copper nitrate. Both answers I got were the same on this: 2.96g of 98% copper nitrate hexahydrate in 1 liter of water.

The second part of the question was how much 70% nitric acid to add to this liter of solution to increase the nitrate ion concentration to five times normal.  Here are the two answers I got.

ANSWER #1 ===============================
"For the nitrate ion concentration to be 5 times greater, add HNO3 (70%) in the ratio of 3.2 mL per L of solution. This is based on the fact that 70% HNO3 is 15.6 molar. A 1M solution requires about 64mL of the 70% HNO3 stock solution per liter of diluted solution. If you want it to be 0.05M, then the volume becomes 5 x 0.64mL, or 3.2 mL of 70% HNO3 per liter of solution."

ANSWER #2 ===============================
"For 1L of a 1M copper nitrate solution, there are 0.02 moles of nitrate ion. If you want a 'nitrate ion concentration five times that of the 0.01 mol/L solution' or a solution with 0.1 moles of nitrate ion (0.02 moles nitrate ion * 5). You already have 0.02 moles of nitrate ion, so you only need an additional 0.08 moles of nitrate ion. 70% nitric acid (HNO3) is 14.6 M. You need 0.08 moles of nitrate ion from our 14.6M solution. 1 mole of nitric acid contains 1 mole of nitrate ion. So, [0.08 moles nitrate ion] / [14.6 mol/L] = 0.005479 L (or 5.479 mL) of 70% nitric acid."

So, you see, both answers were thoughtfully presented, including all the math, but one came out to 3.2mL and the other 5.479mL.  Again, I'm not in a position to argue with either one (having barely passed high school Chemistry more than 50 years ago!) but would like to understand the proper derivation and get the solution (pardon the pun) right.

Many thanks... anyone!

[Borek]

Second one is closer, but slightly off.

The difference comes from the fact that first assumes initial concentration of nitrate ion to be 0.01M, while it is 0.02M, and it ignores the initial concentration. First recipe would be correct if you were planning to make a 0.05M solution adding nitric acid to pure water. That's not what you will be doing.

I got 5.2 mL using concentration calculator. Why not 5.5 mL? Because stock solutions of nitric acid are 69% w/w and 15.4 M (not 14.6 M).

[Inojim]

Thanks, Borek;

That's the confirmation I needed.  I appreciate your explanation... sounds right to me.