Calculate the concentrations of all species present in a 0.33 M solution of ethylammonium chloride (C2H5NH3Cl).
Okay so i tried doing
HA = H + A
ice table regular it doesnt work
idk what Ka to use. b/c of the chloride.
i have to find [C2H5NH2][H+][OH -][C2H5NH3+][Cl -]
C2H5NH2 is a base but you don't have the base. You have the salt, C2H5NH3Cl which contains the conjugate acid of the base or C2H5NH3^+. It's the acid that is hydrolyzing (reacting) with the water to give the base. You can look up Kb but no tables (at least none I know of) are around for the Ka. So we calculate Ka by Kw/Kb since we CAN look up Kb for C2H5NH2
No, I think if you will read my former response closely that isn't what I said to do.
C2H5NH3^+ + HOH ==> C2H5NH2 + H3O^+
Ka = Kw/Kb = (C2H5NH2)(H3O^+)/(C2H5NH3^+)
Now plug in Kw for water, Kb for ethylamine, 0.33 for the salt and calculate the y and y in the numerator. I get approximately 3 x 10^-6 M for H3O^+ so it should be close to that. I don't know what your value is for Kb but I looked up ethylamine on the internet and found 4.7 x 10^-4. Obviously you should use your book's value.
This is a salt so it dissolves (and hydrolyzes in H2O).
C2H5HN3Cl(aq)==> C2H4NH3^+ + Cl^-
That gives you the Cl^- immediately. It's 0.33 M.
Now the salt dissolves in water (use NH4^+ as an example).
C2H5NH3^ + HOH ==> C2H4NH2 + H3O^+
Does that help? Use Ka = Kw/Kb and Kb is for C2H4NH2 (ethylamine).
Well, let's start unraveling this mystery.
First, let's label the given compounds to make things easier:
HA: C2H5NH3Cl
H+: H+
A-: Cl-
Now, you mentioned something about an ice table, but we're dealing with a strong base here, so we can skip that and focus on the dissociation.
The equation for the dissociation of ethylammonium chloride is:
C2H5NH3Cl → C2H5NH3+ + Cl-
Since ethylammonium chloride completely dissociates in water, we can say that the initial concentration of C2H5NH3Cl is 0.33 M.
However, as we mentioned, this is a strong base. Therefore, we can assume that most of it will dissociate completely.
After dissociation, we have:
[H+] = [C2H5NH3+] = 0.33 M (since they are both the conjugate acid of the base)
[OH-] = 0 (since this is an acid and doesn't produce hydroxide ions)
[Cl-] = 0.33 M (since it comes from the dissociation of ethylammonium chloride)
[C2H5NH2] = 0 M (since this is the base, and it doesn't exist anymore after the dissociation)
So, the concentrations of all species present in the 0.33 M solution of ethylammonium chloride are:
[H+] = [C2H5NH3+] = 0.33 M
[OH-] = 0
[Cl-] = 0.33 M
[C2H5NH2] = 0 M
I hope that clears things up for you. If not, don't hesitate to let me know, and I'll try to bring out the big clown shoes to make it even clearer!
To calculate the concentrations of all species present in a 0.33 M solution of ethylammonium chloride (C2H5NH3Cl), you need to consider the dissociation of the salt in water.
First, let's write the balanced chemical equation for the dissociation of ethylammonium chloride:
C2H5NH3Cl (solute) -> C2H5NH3+ (aqueous) + Cl- (aqueous)
In this dissociation equation, C2H5NH3+ is the conjugate acid of ethylamine (C2H5NH2) and Cl- is the conjugate base.
Now, let's define the variables:
- [C2H5NH3+] represents the concentration of ethylammonium ion.
- [C2H5NH2] represents the concentration of ethylamine.
- [H+] represents the concentration of hydrogen ion.
- [OH-] represents the concentration of hydroxide ion.
- [Cl-] represents the concentration of chloride ion.
Since you have a 0.33 M solution of ethylammonium chloride, the concentration of C2H5NH3Cl is 0.33 M.
Now, based on the balanced equation, we can say that:
[C2H5NH3+] = [Cl-] (assuming complete dissociation)
Thus, [C2H5NH3+] = 0.33 M.
Since ethylamine is a weak base, it will undergo partial ionization in water. To find the concentration of ethylamine ([C2H5NH2]) and hydrogen ions ([H+]), you can consider the equilibrium expression for the ionization of a weak base:
C2H5NH2 (solute) + H2O (solvent) ⇌ C2H5NH3+ (aqueous) + OH- (aqueous)
The equilibrium constant for this ionization can be denoted as Kb.
To find the concentrations, you need to set up an ICE (Initial, Change, Equilibrium) table and use the given Kb value for ethylamine. However, since ethylamine is a weak base and not a strong acid, you cannot directly use the pH equation. Instead, you will need to use the equation for calculating the hydroxide concentration.
The pH equation for calculating the hydroxide concentration (OH-) is:
pOH = -log[OH-]
Once you calculate pOH, you can find the hydroxide concentration ([OH-]) using the equation:
[OH-] = 10^(-pOH)
Remember that in a neutral solution, [H+] = [OH-] = 1.0 x 10^(-7) M.
Given that ethylamine has a Kb value of 6.4 x 10^(-4) at a certain temperature, you will need to use this value to calculate the concentrations of ethylamine ([C2H5NH2]) and [H+].
By setting up the ICE table and using the Kb value, you can determine the equilibrium concentrations of [C2H5NH2], [H+], and [OH-].
To summarize:
1. Use the balanced equation to determine the concentrations of [C2H5NH3+] and [Cl-], which are both equal to 0.33 M.
2. Utilize the Kb value for ethylamine to set up an ICE table and calculate the equilibrium concentrations of [C2H5NH2], [H+], and [OH-].
3. Finally, use the equilibrium concentrations to deduce all the concentrations you need: [C2H5NH2], [H+], [OH-], and [Cl-].
Keep in mind that this process assumes complete dissociation of the salt, which is not strictly valid for concentrated solutions. Nevertheless, it can provide approximate concentrations in dilute solutions.
I don't understand why u did the Ka=Kw/Kb
why cant u just use Kb? isnt it a base?
no because i keep getting .01359 for C2H4NH3^+
but obviously that's wrong
i did x^2/.33-x=5.6e-4