You initially have 3 moles of CH3COOH (HA) and 3 moles CH3COO- (A-). When you add 0.01 mol of HCl, that will dissociate to give you 0.01 mol of H+ ions. That 0.01 mol of H+ ions will react with 0.01 mol of CH3COO- to form 0.01 mol of CH3COOH. You now have 3.01 mol CH3COOH and 2.99 mol CH3CHOO-
Wow. This lesson made headache. I am still confused and tomorrow is my exam. This acid and bases and buffer were the only who I didn’t get it but trying my best to understand it.
At 7:00, can you explain why the [A-] equals 2.99? What do you mean by 0.01 mol gets consumed by the hydrogen ion concentration?
You initially have 3 moles of CH3COOH (HA) and 3 moles CH3COO- (A-). When you add 0.01 mol of HCl, that will dissociate to give you 0.01 mol of H+ ions. That 0.01 mol of H+ ions will react with 0.01 mol of CH3COO- to form 0.01 mol of CH3COOH. You now have 3.01 mol CH3COOH and 2.99 mol CH3CHOO-
Hi, why don't we use 3 for molarity of the conjugate base? Doesn't the acid also dissociate into H+ and A-?
Wow thank you very much 🔥🔥
You're welcome 😊
Wow. This lesson made headache. I am still confused and tomorrow is my exam. This acid and bases and buffer were the only who I didn’t get it but trying my best to understand it.
🙌 Thank you!
Sorry, just to clarify - for the second example, does [A-] refer to NaCH3COO and [HA] refer to CH3COOH?
Yes, [A-] is for CH3COO- and [HA] is for CH3COOH!
brain explosionnnnnn