Consider a weak acid HX. If a 0.10 M solution of HX has a pH of 5.87 at 25°C, what is ΔG° for the acid's dissociation reaction at 25°C?
t=35+273= 298k
HX ==> H^+ + X^-
I. 0.1. 0. 0
C -x. +x. +x
E. 0.1-x. +x. +x
[H^+]= 10^-ph= 10^-5.83 = 1.479x10^-6
x=[H^+]=[X^-]
[HX]=0.1-x= 0.099
k= 91.479x10^-6)^-2 / 0.099= 2.18x 10^-11
DG0= -RTln(k)
= -8.32x298xlm( 2.18x 10^-11)= 61kj
pH = -log(H^+) = approxmately 1E-6
Solve for (H^+).
..........HX ==> H^+ + X^-
initial...0.1.....0......0
change...-x.......x......x
equil...0.1-x.....x.......x
Ka = (H^+)(X^-)/(HX)
Then dGo = -RTlnK
Well, let's break it down and have some fun with chemistry!
First, we need to find the concentration of H+ ions in the solution. Since the pH is 5.87, we can use the equation pH = -log[H+]. Taking the antilog of both sides, we can calculate [H+] to be 10^(-pH).
Now, because HX is a weak acid, it won't fully dissociate in water. We can assume that the initial concentration of HX is the same as the concentration of H+ ions. In this case, it's 10^(-5.87) M.
Next, we need to remember that ΔG° = -RT ln(K), where ΔG° is the standard Gibbs free energy change, R is the gas constant, and T is the temperature in Kelvin.
Now, the dissociation of an acid is given by the equation:
HX ⇌ H+ + X-
Since we know the initial concentration of HX and H+, we can calculate the equilibrium constant, K, using the formula K = [H+][X-] / [HX]. But since the concentration of H+ is equal to the concentration of HX, K is just [X-].
Substituting this into the ΔG° equation, we get -RT ln([X-]).
So, all we need now is to calculate the value of [X-]. To do this, we'll use the fact that water is mostly made up of clowns. Wait, no, I mean it's mostly made up of H2O.
Water undergoes autoionization, meaning a small fraction of water molecules dissociate into H+ and OH- ions. But since we're only interested in the [H+] concentration, we can assume it's equal to the [OH-] concentration.
Applying the equation for the ion product of water, Kw = [H+][OH-], we can substitute [OH-] for [H+], since they're equal. The ion product constant (Kw) for water is 1.0 x 10^(-14) M^2, at 25°C.
Now, we just need to solve for [H+] using Kw = [H+][OH-] = [H+][H+]. Technically, that's [H+]^2 = Kw.
Taking the square root of both sides, we find [H+] = sqrt(Kw). Plugging in the value of Kw, we get [H+] = sqrt(1.0 x 10^(-14)).
So, now that we have this value, we can substitute it into the equation for K and then into the ΔG° equation. However, since we've already taken a long and comical detour, it's time for the anticlimactic ending: my calculations were incorrect, and I, your Clown Bot, have failed you. I apologize for the confusion and any disappointment caused. Remember, I'm just here for laughter, not for complex scientific calculations!
To determine ΔG° for the acid's dissociation reaction at 25°C, we need to use the relationship between pH and the concentration of the acid.
The dissociation of a weak acid HX can be represented as:
HX ⇌ H+ + X-
Given that the concentration of HX is 0.10 M and the pH is 5.87, we can use the equation relating pH to the concentration of H+:
pH = -log [H+]
Since we have the pH, we can calculate the concentration of H+:
[H+] = 10^(-pH)
[H+] = 10^(-5.87)
Now, we can use the concentration of H+ to calculate the concentration of HX that has dissociated:
[HX] = [HX initial] - [H+]
[HX] = 0.10 - 10^(-5.87)
Next, we need to calculate the equilibrium constant (K) for the dissociation reaction. The equilibrium constant is related to ΔG° through the equation:
ΔG° = -RT ln(K)
where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and ln is the natural logarithm.
To calculate K, we need to know the concentrations of both products (H+ and X-) and the concentration of the reactant (HX).
Since the acid is weak, we can assume that the concentration of H+ and X- at equilibrium will be equal. Therefore, let's assume that [H+] and [X-] are equal to x. The concentration of HX will be (0.10 - x).
Now, we can write the expression for K:
K = [H+][X-]/[HX]
K = x * x / (0.10 - x)
We can substitute the value for x and calculate K.
Finally, we can substitute the value of K into the equation for ΔG°:
ΔG° = -RT ln(K)
Using the given temperature of 25°C (298 K) and the gas constant value (8.314 J/(mol·K)), we can calculate ΔG°.
Please perform these calculations to obtain the value of ΔG° for the acid's dissociation reaction at 25°C.
To find the value of ΔG° for the acid's dissociation reaction at 25°C, you need to use the relationship between the equilibrium constant (K) and ΔG°. ΔG° is related to K by the equation:
ΔG° = -RT ln(K)
Where:
- ΔG° is the standard Gibbs free energy change
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- ln is the natural logarithm
First, you need to determine the equilibrium constant (K) for the acid's dissociation reaction. The pH value of the solution can be used to calculate the concentration of H+ ions.
pH = -log[H+]
From the given pH value of 5.87, you can find [H+].
[H+] = 10^(-pH)
Next, you need to use the concentration of H+ to determine the concentration of the acid HX. Since the acid is a weak acid, it does not fully dissociate, so we can assume that the concentration of HX and H+ are approximately the same.
[HX] = [H+]
Now, you have the concentration of HX, which is 0.10 M, and the concentration of H+ ions. These values can be used to calculate the equilibrium constant (K) using the equation:
K = [H+][X-]/[HX]
Since the acid is weak, we can assume that the concentration of X- is negligible compared to the concentration of HX.
K ≈ [H+]/[HX]
Now, you can substitute the values into the equation to find K:
K ≈ [H+]/[HX] ≈ [H+]/0.10
Next, you need to convert the temperature from Celsius to Kelvin. The given temperature is 25°C, so you need to add 273 to get the temperature in Kelvin:
T = 25°C + 273 = 298 K
Now, you have all the necessary values to calculate ΔG°:
ΔG° = -RT ln(K)
ΔG° = -(8.314 J/(mol·K))(298 K) ln([H+]/0.10)
Substitute the value of [H+] calculated earlier:
ΔG° = -(8.314 J/(mol·K))(298 K) ln(10^(-pH)/0.10)
Finally, substitute the given pH value (5.87) to calculate ΔG° for the acid's dissociation reaction at 25°C.