# 1. Calculate the concentration of H3O+, HC2O4-1, and oxalate ion (C2O42-) in a 0.175 M solution of oxalic acid (C2H2O4).

[For oxalic acid, Ka1 = 6.5 ´ 10-2, Ka2 = 6.1 ´ 10-5.]

## To calculate the concentrations of H3O+, HC2O4-1, and oxalate ion (C2O42-) in a 0.175 M solution of oxalic acid (C2H2O4), we need to determine the dissociation of oxalic acid and use the given Ka values.

The chemical equation for the dissociation of oxalic acid is:

C2H2O4 ⇌ H+ + HC2O4-

According to the equation, one mole of oxalic acid dissociates to produce one mole of H+ and one mole of HC2O4-. Therefore, the initial concentration of H+ and HC2O4- will be the same.

Since the initial concentration of oxalic acid is 0.175 M, the initial concentration of H+ and HC2O4- will also be 0.175 M.

For the second dissociation, the equation is:

HC2O4- ⇌ H+ + C2O42-

The equilibrium concentrations of H+ and C2O42- can be calculated using the Ka2 value (6.1 ´ 10-5) and the initial concentration of HC2O4-.

Let's denote the concentration of H+ and C2O42- as [H+] and [C2O42-], respectively. The initial concentration of HC2O4- is 0.175 M.

Using the Ka2 expression for the dissociation of HC2O4-:

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

Rearranging the equation:

[H+][C2O42-] = Ka2 × [HC2O4-]

Plugging in the values:

[H+][C2O42-] = (6.1 ´ 10-5) × (0.175 M)

Solving the equation gives us the concentrations of [H+] and [C2O42-].

Therefore, the concentration of H3O+ is 0.175 M, the concentration of HC2O4- is 0.175 M, and the concentration of oxalate ion (C2O42-) is (6.1 ´ 10-5) × (0.175 M).

## To calculate the concentration of H3O+, HC2O4-1, and oxalate ion (C2O42-) in a 0.175 M solution of oxalic acid (C2H2O4), you will need to consider the acid dissociation reactions and the equilibrium constants.

Oxalic acid (C2H2O4) dissociates in two steps:

1. The first dissociation reaction is:

C2H2O4 ⇌ HC2O4-1 + H+

The equilibrium constant (Ka1) for the first dissociation reaction is given as 6.5 x 10^-2.

2. The second dissociation reaction is:

HC2O4-1 ⇌ C2O42- + H+

The equilibrium constant (Ka2) for the second dissociation reaction is given as 6.1 x 10^-5.

To calculate the concentration of each species, you will need to set up equations based on these dissociation reactions and the given equilibrium constants.

Let's define x1 as the concentration of HC2O4-1, x2 as the concentration of C2O42-, and x3 as the concentration of H3O+.

For the first dissociation reaction:

HC2O4-1 + H2O ⇌ H3O+ + C2H2O4

Using the given initial concentration of oxalic acid (C2H2O4) as 0.175 M, we can assume that the initial concentration of HC2O4-1 is also 0.175 M.

Using the equation:

K1 = [H3O+][C2H2O4]/[HC2O4-1]

Plugging in the values:

6.5 x 10^-2 = [x3][0.175]/[0.175]

Since the concentration of HC2O4-1 is equal to the concentration of C2H2O4, we can substitute [C2H2O4] for [HC2O4-1]:

6.5 x 10^-2 = [x3][0.175]/[0.175]

Simplifying the equation, we get:

[x3] = (6.5 x 10^-2)[0.175]

So the concentration of H3O+ is (6.5 x 10^-2)(0.175) or 1.1375 x 10^-2 M.

Now let's move on to the second dissociation reaction:

C2O42- + H2O ⇌ C2H2O4 + OH-

Using the equation:

K2 = [C2H2O4][OH-]/[C2O42-]

Plugging in the values:

6.1 x 10^-5 = [0.175][OH-]/[x2]

Since the concentration of C2H2O4 is equal to the initial concentration of oxalic acid, we can substitute [C2H2O4] for 0.175:

6.1 x 10^-5 = [0.175][OH-]/[x2]

Simplifying the equation, we get:

[OH-] = (6.1 x 10^-5)(0.175)/[x2]

Now, since water (H2O) dissociates to generate equal concentrations of H3O+ and OH-, we can assume that [OH-] is equal to [x3]. Therefore, we can substitute [x3] for [OH-] in the equation:

1.1375 x 10^-2 = (6.1 x 10^-5)(0.175)/[x2]

Simplifying the equation, we get:

[x2] = (6.1 x 10^-5)(0.175)/[1.1375 x 10^-2]

Therefore, the concentration of C2O42- is (6.1 x 10^-5)(0.175)/(1.1375 x 10^-2) or 9.3462 x 10^-4 M.

To find the concentration of HC2O4-1, we can use the equation:

[HC2O4-1] = [C2H2O4] - [H3O+]

Since the initial concentration of oxalic acid is 0.175 M, and we have already calculated the concentration of H3O+ as 1.1375 x 10^-2 M, we can substitute these values:

[HC2O4-1] = 0.175 - 1.1375 x 10^-2

Therefore, the concentration of HC2O4-1 is 0.175 - 1.1375 x 10^-2 or 0.1643 M.

In summary:

- The concentration of H3O+ is 1.1375 x 10^-2 M.

- The concentration of HC2O4-1 is 0.1643 M.

- The concentration of C2O42- is 9.3462 x 10^-4 M.

## .......H2C2O4 + H2O --> H3O^+ + HC2O4^-

I......0.175.............0.......0

C........-x..............x.......x

E.....0.175-x............x.......x

Substitute the E line into Ka1 expression and solve for x = (H3O^+) = (HC2O4^-).

That gives you H3O^+ and HCO4^- and note they are equal. It also gives (H2C2O4). That leaves C2O4^2-

.......HC2O4^- + H2O ==> H3O^+ + C2O4^2-

Ka2 = (H3O^+)(C2O4^2-)/(HC2O4^-)

Since (H3O^+) = (HC2O4^-), then (C2O4^2-) = ka2.