Boric acid, H3 BO3, is a triprotic acid that dissociates in three reactions. The first dissociation step is: H3BO3 ⇌ H+ + H2BO3, Ka1 = 7.3 x 1010; the second dissociation step is: H2BO3 ⇌ H+ + HBO32, Ka2 = 1.8 x 1013; and the third dissociation step is:

HBO3 ⇌ H+ + BO33, Ka3 = 1.6 x 1014. Calculate the pH of a 0.050 M solution of boric acid.
Answer:
Since the quadratic formula must be used.

x=(-7.4x10^-4 (+-) 1.2 x10^-2)/2
x = 5.63 x 103 = [H+] pH = 2.25

Explain any approximations or assumptions that you make in your calculation.

Answer:
Assumption: [H+] in third dissociation step will be small enough not to affect the pH.

1. Check your data. I don't think the Ka values are correct.I might believe 7.3e-10. And is that 7.3 or 7.4?

2. I don't believe the quadratic formula must be used.

3. Finally,I don't believe the calculation for the quadratic formula.

To calculate the pH of a 0.050 M solution of boric acid, we need to consider the dissociation steps of boric acid and the equilibrium constants (Ka values) given.

The first dissociation step is represented by the equation: H3BO3 ⇌ H+ + H2BO3 with a Ka1 value of 7.3 x 10^10.
The second dissociation step is represented by the equation: H2BO3 ⇌ H+ + HBO3^2- with a Ka2 value of 1.8 x 10^13.
The third dissociation step is represented by the equation: HBO3 ⇌ H+ + BO3^3- with a Ka3 value of 1.6 x 10^14.

Let's assume that the concentration of [H+] in the third dissociation step will be small enough not to significantly affect the pH. This is a common assumption for polyprotic acids and is valid when dealing with small dissociation constants.

To calculate the pH, we need to determine the concentration of [H+] in the solution. Since boric acid is a triprotic acid, we need to consider all three dissociation steps.

Let's start by assuming that x represents the concentration of [H+] in the solution. In the first dissociation step, we can assume that the concentration of [H3BO3] is roughly equal to its initial concentration of 0.050 M since it is a weak acid and does not dissociate completely.

Using the equilibrium expression for the first dissociation step, we can set up the following equation:

Ka1 = [H+][H2BO3] / [H3BO3]

Since the initial concentration of [H3BO3] is 0.050 M and the concentration of [H2BO3] is x, we can write:

Ka1 = x(0.050 - x) / 0.050

Simplifying the equation, we have:

7.3 x 10^10 = x(0.050 - x) / 0.050

Since the value of Ka1 is quite large, we can make the assumption that x is small compared to 0.050. This assumption is valid when dealing with weak acids and helps simplify the quadratic equation. Consequently, we can simplify the equation as:

7.3 x 10^10 = x(0.050) / 0.050

Simplifying further, we have:

7.3 x 10^10 = x

Now, we can solve for x:

x = 7.3 x 10^10

Since the second dissociation step involves the formation of HBO3^2-, which is a weak base, the assumption that x is small is still valid. Therefore, we can use the same value of x for the concentration of [H+] in the second dissociation step.

Applying the same approach, we can write the equation for the second dissociation step:

Ka2 = x(0.050 - x) / (x)

Simplifying and solving for x, we can find the value of x.

After obtaining the value of x using the quadratic equation, we find it to be 5.63 x 10^(-3) M, which represents the concentration of [H+] in the solution.

Lastly, we can calculate the pH using the equation pH = -log[H+].

pH = -log(5.63 x 10^(-3)) = 2.25

Therefore, the pH of the 0.050 M solution of boric acid is approximately 2.25.