ok the buffer solution we made contained 3.3g of sodium acetate trihydrate, 46mL

Question

ok the buffer solution we made contained 3.3g of sodium acetate trihydrate, 46mL of DI water, and 4.0mL of 6.0M HC2H3O2. From there we only took a 19mL sample and now I have to find how many moles of sodium acetate are contained in the 19ml buffer then I need to find how many moles of HCL would need to be added to the 19ml buffer to destroy the system (ie. neutralize the base) and then I need to find how many mL of a 3.0 M HCl would be needed to destroy 19mL of the buffer system. I have no idea on how to go about solving this so if anyone can give me some help I would really appreciate it.

Explanation / Answer

Destroying with HCl means reacting all sodium acetate. 3.3g=0,024 moles sodium acetate. We will assume that the final buffering solution have 50mL. Therefore 19 mL involve 19/50*0.024=0.009 moles. we need 0.009 moleof HCl. May I approximate to 0.01 mole? Thank you. Now, HCl is at your choice. Pure HCl is gaseous, then we need anaqeuous solution. The most concentrated solution is around 36%, but I believe we willuse more appropriate 0.1N (or 0.1M) solution. The amount will bearound 100 mL. Hope no math mistakes.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.