I just got an email from a guy that bought 35% nitric and wanted to know how much it would take to dissolve a gram of silver. When I answered him, like most everyone else, I just assumed that 35% would be the same as diluting 70% with an equal amount of water. In thinking about this further, I find this to be in error. Actually, when diluting 70% nitric, 50/50, the result is close to 41%, not 35%.
http://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry
Using this chart, here is the math:
What happens when you dilute 70%, by weight, nitric 50/50? The specific gravity (g/ml or g/cc), at 20C, of 70% is 1.4134 (see chart). Therefore, a liter of 70% would weigh 1413.4 grams. Therefore, a liter of 70% would contain 1413.4 x .70 = 989.4 grams of nitric. By adding a liter of water, which weighs about 1000 grams, to a liter of 70% nitric (thereby diluting it 50/50), the total weight would be 1413.4 + 1000 = 2413.4 grams. Therefore, the final acid strength, by weight, of 50/50 70% nitric, would be 989.4/2413.4 = .41 or 41%.
Here again, by the chart, a liter of 41% nitric would contain 1252.7 x .41 = 513.6 grams of nitric. On the other hand, by the chart, a liter of 35% nitric would only contain 1214 x .35 = 424.9 grams of nitric. Therefore, the man I advised would require 513.6/424.9 = 1.209 times more 35% nitric than I told him. I told him to use 2.44 ml of 35%/gram of silver. It would actually take 2.44 x 1.209 = 2.95 ml of 35% nitric to dissolve one gram of silver.
As you can see, the math on this is not intuitive. You have to use a percentage/specific gravity chart like the one above in order to get the correct answer. There are other ways, but this is the easiest.
This works with all acids or any other solution that you can find a specific gravity chart on.
I did this math hurriedly. If anyone finds an error in my logic or my math, please let me know.
I might note that all this can vary slightly, depending on the actual temperatures involved. I used the specific gravities at 20C (68F) for all calculations.
Chris
http://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry
Using this chart, here is the math:
What happens when you dilute 70%, by weight, nitric 50/50? The specific gravity (g/ml or g/cc), at 20C, of 70% is 1.4134 (see chart). Therefore, a liter of 70% would weigh 1413.4 grams. Therefore, a liter of 70% would contain 1413.4 x .70 = 989.4 grams of nitric. By adding a liter of water, which weighs about 1000 grams, to a liter of 70% nitric (thereby diluting it 50/50), the total weight would be 1413.4 + 1000 = 2413.4 grams. Therefore, the final acid strength, by weight, of 50/50 70% nitric, would be 989.4/2413.4 = .41 or 41%.
Here again, by the chart, a liter of 41% nitric would contain 1252.7 x .41 = 513.6 grams of nitric. On the other hand, by the chart, a liter of 35% nitric would only contain 1214 x .35 = 424.9 grams of nitric. Therefore, the man I advised would require 513.6/424.9 = 1.209 times more 35% nitric than I told him. I told him to use 2.44 ml of 35%/gram of silver. It would actually take 2.44 x 1.209 = 2.95 ml of 35% nitric to dissolve one gram of silver.
As you can see, the math on this is not intuitive. You have to use a percentage/specific gravity chart like the one above in order to get the correct answer. There are other ways, but this is the easiest.
This works with all acids or any other solution that you can find a specific gravity chart on.
I did this math hurriedly. If anyone finds an error in my logic or my math, please let me know.
I might note that all this can vary slightly, depending on the actual temperatures involved. I used the specific gravities at 20C (68F) for all calculations.
Chris
