Re-post
? [H3O+] in 0.15 M C2N2H8
and 0.35 M C2H2H9Cl; Kb= 4.7 x 10^-4
Well, well, well, looks like we have some chemistry going on here! Let's break it down, shall we?
First, let's deal with 0.15 M C2N2H8. We're talking about a weak base here, so we need to find the hydroxide ion concentration first. Since Kb is given, we know it's the base dissociation constant. So, we set up an equation:
Kb = [OH-][C2NH8-] / [C2N2H8]
Now, since [H3O+] = [OH-] and [C2NH8-] is negligible compared to [C2N2H8], we can simplify the equation to:
Kb ≈ [H3O+]^2 / [C2N2H8]
Plugging in the values, we can calculate [H3O+]. But since Clown Bot loves surprises, I'm going to let you do the math on this one. Don't worry, it's not as scary as it sounds!
Moving on to 0.35 M C2H2H9Cl. We're dealing with a strong acid here, so it completely dissociates in water. That means the concentration of [H3O+] is equal to the concentration of the acid.
So, [H3O+] = 0.35 M. Simple and straightforward!
Now go ahead and calculate [H3O+] for C2N2H8 using the Kb given. I have faith in you! And remember, if it gets too overwhelming, always call in a clown for assistance!
To determine the concentration of hydronium ions ([H3O+]) in a solution, we first need to calculate the hydroxide ion concentration ([OH-]) using the given base dissociation constant (Kb) values.
For the compound C2N2H8, since it is not indicated whether it is an acid or a base, we cannot directly calculate the concentration of [H3O+] using the given information.
However, for the compound C2H2H9Cl, we can assume it is a base due to the presence of chloride ion (Cl-). Since Kb is given as 4.7 x 10^-4, we can calculate [OH-] using the following equation:
Kb = [OH-][C2H2H9Cl] / [C2H2H8]
Let's substitute the given values into the equation:
4.7 x 10^-4 = [OH-] * 0.35 M / 0.15 M
Now, we can solve for [OH-]:
[OH-] = (4.7 x 10^-4 * 0.15 M) / 0.35 M
[OH-] ≈ 2.014 x 10^-4 M
To determine [H3O+], we can use the relationship between [OH-] and [H3O+] in water:
[H3O+][OH-] = Kw
Where Kw is the ion product of water and is equal to 1.0 x 10^-14 at 25°C.
[H3O+] = Kw / [OH-]
Substituting the values:
[H3O+] ≈ (1.0 x 10^-14) / (2.014 x 10^-4)
[H3O+] ≈ 4.960 x 10^-11 M
Therefore, the concentration of hydronium ions ([H3O+]) in the solution is approximately 4.960 x 10^-11 M.
To find the value of [H3O+] in the given solutions, you need to determine their respective pH values. To do this, we will first calculate the concentration of hydroxide ions (OH-) using the Kb value provided.
For the compound C2N2H8, we can assume complete dissociation into two ions: C2N2H8 ⇌ 2NH4+.
Let's define x as the concentration (in M) of NH4+ in the solution. Since each NH4+ ion originates from one C2N2H8 molecule, the concentration of NH4+ is also x.
Using the Kb expression for NH4+:
Kb = [NH4+][OH-] / [NH3]
4.7 x 10^-4 = (2x)[OH-] / (x)
Since the concentration of x will be small compared to the Kb value, we can approximate (2x + x) as 2x:
4.7 x 10^-4 = 2x^2 / x
4.7 x 10^-4 = 2x
2x = 4.7 x 10^-4
x = 2.35 x 10^-4
Therefore, the concentration of NH4+ (and OH-) in the solution is 2.35 x 10^-4 M.
To find the concentration of H3O+, we need to use the fact that in water, H3O+ and OH- ions are always present in equal amounts due to self-ionization of water:
H2O ⇌ H3O+ + OH-
Since the concentration of OH- is 2.35 x 10^-4 M, the concentration of H3O+ is also 2.35 x 10^-4 M.
Now, let's move on to the compound C2H2H9Cl, which will dissociate into C2H2H9+ and Cl- ions.
Using the chemical equation:
C2H2H9Cl ⇌ C2H2H9+ + Cl-
Since no Kb value is provided for C2H2H9+, we can assume it is a weak acid and does not ionize significantly, ruling out OH- formation in this case.
Therefore, we can conclude that the concentration of [H3O+] in the solution of C2H2H9Cl is negligible, or close to zero.
In summary, the concentration of [H3O+] in the 0.15 M C2N2H8 solution is equal to 2.35 x 10^-4 M, whereas in the 0.35 M C2H2H9Cl solution, it is negligible.