In a 1.0× 10–2 M solution of CH3NH3Br(aq), identify the relative molar amounts from highest to lowest for :
-Br-
-OH-
-CH3NH3+
-HBr
-CH3NH2
-H3O+
-H2O
Damn Bob you left nothing to help
I can't find the value x by calculation. It ends up to give me a quadratic equation and the value of x is negative! please help me ASAP...
I miss 2013
Well, this sounds like a chemist's playground! Brace yourself for some chemical comedy! Here's the ranking of the relative molar amounts from highest to lowest:
1. HBr: Oh, that's a hot shot! It's like the heavyweight champ of the ring. HBr is a strong acid, so it ionizes almost completely in solution.
2. H3O+: Talk about an H+ party! H3O+ is commonly known as hydronium ion, and it forms when water (H2O) accepts a proton. It's like a water molecule that's had way too many hydrogen shots.
3. Br-: Look who's joining the party! Br- is the bromide ion, and it's negatively charged. Don't underestimate it, though. It might be small, but it packs a punch.
4. -CH3NH3+: Here come the methylammonium ions, dancing their way onto the scene. Don't worry, they're positively charged and ready to mingle.
5. -OH-: Oh, hello there, hydroxide ion! It's like a really dedicated partygoer who can't help but take on an extra electron. Get ready for some alkaline vibes.
6. -CH3NH2: Uh-oh, seems like the party is winding down. Here come the methylamine molecules, looking a bit lonely without their positive charge. Still, they're up for a good time, just not as lively as the ions.
7. H2O: Ah, the unsung hero of the night. H2O is always there, ready to provide hydration and stability. It might not steal the show, but it's the backbone of the entire party.
So, in summary, the ranking from highest to lowest relative molar amounts is: HBr > H3O+ > Br- > -CH3NH3+ > -OH- > -CH3NH2 > H2O. Hope this chemical circus brought a smile to your face!
To identify the relative molar amounts from highest to lowest in a solution of CH3NH3Br(aq), we can look at the dissociation of the compound. CH3NH3Br is an ionic compound that dissociates in water.
The dissociation of CH3NH3Br can be represented by the following equation:
CH3NH3Br(aq) ⇌ CH3NH3+(aq) + Br-(aq)
Based on this dissociation equation, we can determine the relative molar amounts as follows:
1. -CH3NH3+: This is the cation formed by the dissociation of CH3NH3Br. It is present in the highest molar amount because every formula unit of CH3NH3Br that dissociates produces one CH3NH3+ ion.
2. -Br-: This is the anion formed by the dissociation of CH3NH3Br. It is present in the same molar amount as CH3NH3+ because the compound dissociates in a 1:1 ratio.
3. -HBr: HBr is the acid formed when CH3NH3Br is dissolved in water. The concentration of HBr can be calculated using the amount of CH3NH3+ formed. Since the molar amount of CH3NH3+ is equal to the molar amount of Br-, the concentration of HBr is the same as the concentration of Br-. Therefore, the molar amount of HBr is the same as the molar amount of Br-.
4. -H3O+: H3O+ ions are formed when HBr molecules donate a proton (H+) to water molecules. The concentration of H3O+ can be calculated using the concentration of HBr, assuming complete ionization of HBr. Therefore, the molar amount of H3O+ is the same as the molar amount of HBr.
5. -OH-: OH- ions are formed in water through the self-ionization of water. However, in this case, CH3NH3Br is an ionic compound that does not produce OH- ions. Therefore, the molar amount of -OH- is zero.
6. -CH3NH2: CH3NH2 is the parent amine compound before the addition of HBr. In this case, it is not present in the solution because it has been converted to CH3NH3+ (the CH3NH3+ molar amount is equivalent to the CH3NH2 initial molar amount).
Therefore, the relative molar amounts from highest to lowest are:
1. -CH3NH3+
2. -Br-
3. -HBr
4. -H3O+
5. -OH-
6. -CH3NH2
........CH3NH3Br ==> CH3NH3^+ + Br^-
I......0.01...........0..........0
C.....-0.01..........0.01.......0.01
E........0...........0.01.......0.01
Then the hydrolysis of the cation.
......CH3NH3^+ + H2O ==> H3O^+ + CH3NH2
I.....0.01................0........0
C......-x.................x........x
E.....0.01-x..............x........x
Ka for CH3NH3^+ = (Kw/Kb for CH3NH2) = (x)(x)/(0.01-x)
Solve for x which gives you H3O^+ and CH3NH2 as well as CH3NH3^+.
Finally, (H^+)(OH^-) = Kw = 1E-14 which allows you to calculate OH^-
Now you can arrange them in any order you wish.
Post your work if you get stuck.