How do you find the concentration of HCl in 25mL of DI water with .1324 grams of Na2CO3 added to it to keep it constant? This was done by creating a standard solution of approximately .1M solution but now I'm trying to find the exact and am confused with the added Na2CO3. The amount of HCl added to the 25 mL of DI water and .1324 grams of Na2CO3 is 29.066 mL (done by titration)...so where do I go from there to find the concentration?
2007-02-27 10:18:08 · 1 answers · asked by A.I.Disguise 2 in Science & Mathematics ➔ Chemistry
Your question is completely confusing.
How would Na2CO3 keep the concentration of HCl constant?
Do you mean the pH?
Do you want to find the [H+] or the [Cl-]? HCl is a strong acid which dissociates 100% in water and the [H+] will react with [CO3(-2)] making [H+]<[Cl-] so it makes no sense to talk about [HCl].
You say you added 29.066 mL of HCl. What concentration was that? If you know that and it is M1, then in your solution which has 29+25=54 mL volume (approximately) you have
[Cl-]=M1*29.066/54.
2007-02-28 01:25:27 · answer #1 · answered by bellerophon 6 · 0⤊ 0⤋