The following reaction has a Delta G

value of 42.6 kJ/mol at 25oC:
HB(aq) +H2O(l) --> H3O+(aq) +B-(aq)
Calculate Ka for the acid HB.

dG = -RTln*Ka

Solve for Ka
dG = 42600 J
R = 8.314
T = 298

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To calculate the Ka (acid dissociation constant) for the acid HB, we need to use the relationship between ΔG, Ka, and the equilibrium constant Keq.

The standard free energy change ΔG° for a reaction is related to the equilibrium constant Keq as follows:

ΔG° = -RT ln(Keq)

Where:
ΔG° is the standard free energy change (in Joules/mol),
R is the gas constant (8.314 J/(mol K)),
T is the temperature in Kelvin,
ln is the natural logarithm.

First, let's convert the temperature from Celsius to Kelvin:

T = 25°C + 273.15 = 298.15 K

Let's convert ΔG from kJ/mol to J/mol:

ΔG = 42.6 kJ/mol × 1000 J/1 kJ = 42,600 J/mol

Now, we can rearrange the equation to solve for Keq:

ΔG° = -RT ln(Keq)

ln(Keq) = -ΔG° / RT

ln(Keq) = -42,600 J/mol / (8.314 J/(mol K) × 298.15 K)

Now, we can solve for ln(Keq):

ln(Keq) = -42,600 J/mol / (8.314 J/(mol K) × 298.15 K)

ln(Keq) ≈ -18.698

Next, we can solve for Keq by taking the exponential of both sides:

Keq = e^(-18.698)

Keq ≈ 5.132 × 10^(-9)

Since the reaction is given by:
HB(aq) + H2O(l) ↔ H3O+(aq) + B-(aq)

According to the reaction stoichiometry, the concentration of B- is equal to the concentration of HB, as 1 mol of HB reacts to form 1 mol of B-. Therefore, the concentration of HB is equal to the concentration of B-.

To calculate Ka, which is the equilibrium constant for the dissociation of the acid HB, we need to consider the concentration of H3O+ and HB:

Ka = [H3O+][B-] / [HB]

Since the concentrations of H3O+ are negligible compared to the initial concentration of HB, we can assume that [H3O+] ≈ 0.

Ka = [B-] / [HB]

Since the concentration of B- (and HB) is equal to the initial concentration of HB, we can simplify the equation further:

Ka = [HB] / [HB]

Therefore, Ka ≈ 5.132 × 10^(-9)

So, the Ka value for the acid HB is approximately 5.132 × 10^(-9).

To calculate Ka for the acid HB, we need to use the relationship between Delta G and Ka. The formula for Delta G and Ka is as follows:

Delta G = -RT * ln(Ka)

Where:
Delta G is the change in Gibbs free energy
R is the gas constant (8.314 J/(mol*K) or 0.008314 kJ/(mol*K))
T is the temperature in Kelvin
Ka is the acid dissociation constant

Given the value of Delta G (42.6 kJ/mol) and the temperature (25°C), we need to convert the temperature to Kelvin.

T(K) = T(°C) + 273.15
T(K) = 25 + 273.15
T(K) = 298.15 K

Now, we can substitute the values into the equation and solve for Ka:

42.6 kJ/mol = - (0.008314 kJ/(mol*K)) * ln(Ka)

To isolate ln(Ka), we divide both sides by -RT:

ln(Ka) = 42.6 kJ/mol / (0.008314 kJ/(mol*K))

ln(Ka) ≈ 5135.47

Now, we can calculate Ka by taking the exponent of both sides:

Ka = e^(ln(Ka))
Ka ≈ e^5135.47

Using a calculator, we find that Ka is an extremely large number since e^5135.47 is beyond the range of most calculators.

Therefore, the value of Ka for the acid HB is approximately equal to e^5135.47, which can be considered an extremely large number.

3.41e-8