How do we order the folllwing considering how easily they grab H^+ ions, when dissolved in liquid ethanoic acid?
1)OH^-
2)F^-
3)Cl^-
4)Br^-
5)I^-
I think the correct order is 5<4<3<2<1,because I^- is strong acid cation and may not like to grab H^+ ions etc.Am I correct?
I can't help with this as my knowledge of using acetic acid as a solvent is limited.
Never mind and thank you!
To determine the order of how easily each ion grabs H+ ions when dissolved in liquid ethanoic acid, we need to consider the strength of the conjugate acid of each ion. The stronger the conjugate acid, the weaker the respective base.
Here's how you can approach this problem:
1) Start by considering the acidity of the conjugate acids of each ion. The stronger the acid, the weaker the corresponding base. In this case, the conjugate acid of each ion is formed when they grab H+ ions.
- Iodide ion (I-) has a weak conjugate acid, HI, making it a weak base and a strong acid cation.
- Bromide ion (Br-) has a slightly stronger conjugate acid, HBr, making it a slightly stronger base than I-.
- Chloride ion (Cl-) has a stronger conjugate acid, HCl, making it a stronger base than Br- and I-.
- Fluoride ion (F-) has an even stronger conjugate acid, HF, making it a stronger base than Cl-, Br-, and I-.
- Hydroxide ion (OH-) has the strongest conjugate acid, H2O, making it the strongest base of them all.
2) Based on the strength of their conjugate acids, we can conclude that the correct order of how easily these ions grab H+ ions in ethanoic acid is:
1) OH^- (hydroxide ion)
2) F^- (fluoride ion)
3) Cl^- (chloride ion)
4) Br^- (bromide ion)
5) I^- (iodide ion)
So, your initial order was incorrect. The iodide ion (I^-) is the weakest base, and the hydroxide ion (OH^-) is the strongest base in this case.