Question

What is the ph and the oxalate concentration of a 1. 0 m solution of oxalic acid at 25 oc? what is the concentration, in m, of other species in the equilibrium? oxalic acid is a diprotic acid with dissociation constants ka,1

Answer 1

To calculate the pH and oxalate concentration of a 1.0 M solution of oxalic acid, we use the given pKa1 and pKa2 values of 1.25 and 3.81, respectively. Using these values, we can determine the concentrations of H3O+ and oxalate in the solution at 25°C. Additionally, we can calculate the concentrations of other species in equilibrium by using the dissociation equations for oxalic acid.

Step-by-step explanation:

Oxalic acid is a diprotic acid with dissociation constants Ka1 and Ka2. To calculate the pH and oxalate concentration of a 1.0 M solution of oxalic acid, we use the given pKa1 and pKa2 values of 1.25 and 3.81, respectively. Using these values, we can determine the concentrations of H3O+ and oxalate in the solution at 25°C:

1. Calculate the concentration of H3O+ using the equation: [H3O+] = √(Ka1 * [oxalic acid])
2. Calculate the concentration of oxalate using the equation: [oxalate] = [oxalic acid] - [H3O+]

Additionally, we can calculate the concentrations of other species in equilibrium by using the dissociation equations for oxalic acid.

Answer 2

Step-by-step explanation:

To calculate the pH and oxalate concentration of a 1.0 M solution of oxalic acid at 25°C, we need to use the dissociation constants (Ka) of oxalic acid. Oxalic acid is a diprotic acid, which means it can donate two protons in solution.

The dissociation constants for oxalic acid are:

Ka1 = 5.90 × 10^-2

Ka2 = 5.90 × 10^-5

To calculate the pH of a 1.0 M solution of oxalic acid, we need to first determine the concentration of hydrogen ions (H+) in solution. This can be done by considering the equilibrium reactions:

H2C2O4 ⇌ H+ + HC2O4-

HC2O4- ⇌ H+ + C2O42-

The equilibrium expressions for these reactions are:

Ka1 = [H+][HC2O4-]/[H2C2O4]

Ka2 = [H+][C2O42-]/[HC2O4-]

Since the concentration of oxalic acid is 1.0 M, the concentration of HC2O4- and C2O42- can be assumed to be negligible compared to the initial concentration of oxalic acid. Therefore, we can simplify the equilibrium expressions to:

Ka1 = [H+][HC2O4-]/1.0 M

Ka2 = [H+][C2O42-]/[HC2O4-]

Rearranging these equations and solving for [H+], we get:

[H+] = √(Ka1Ka2)/(1.0 M)

Plugging in the values for Ka1 and Ka2, we get:

[H+] = √(5.90 × 10^-2 × 5.90 × 10^-5)/(1.0 M) = 1.08 × 10^-3 M

Using the equation pH = -log[H+], we can calculate the pH of the solution:

pH = -log(1.08 × 10^-3) = 2.97

To calculate the concentration of oxalate ions (C2O42-) in solution, we can use the equilibrium expression for the second dissociation of oxalic acid:

Ka2 = [H+][C2O42-]/[HC2O4-]

We already know the concentration of hydrogen ions ([H+]) from the previous calculation, and we can assume that the concentration of HC2O4- is equal to the initial concentration of oxalic acid (1.0 M). Therefore, we can solve for [C2O42-]:

[C2O42-] = (Ka2 × [HC2O4-])/[H+]

Plugging in the values for Ka2, [HC2O4-], and [H+], we get:

[C2O42-] = (5.90 × 10^-5 × 1.0 M)/(1.08 × 10^-3 M) = 5.46 × 10^-3 M

Therefore, the concentration of oxalate ions in solution is 5.46 × 10^-3 M.

To calculate the concentrations of other species in the equilibrium, we can use the equilibrium expressions for each of the dissociation reactions and the conservation of mass balance:

[H2C2O4] = [H+] + [HC2O4-]

[HC2O4-] = [H2C2O4]/(1 + Ka1/[H+])

[C2O42-] = [HC2O4-] × Ka2/[H+]

Pl