1) Boric acid, H3 BO3, is a triprotic acid that dissociates in three reactions. The first dissociation step is: H3BO3 ⇌ H^+ + H2BO3^-, K(a1) = 7.3 x 10^-10; the second dissociation step is: H2BO3^- ⇌ H^+ + HBO3^-2, K(a2) = 1.8 x 10^-13; and the third dissociation step is:
HBO3 ⇌ H^+ + BO3^-3, K(a3) = 1.6 x 10^-14. Calculate the pH of a 0.050 M solution of boric acid.
2) Explain any approximations or assumptions that you make in your calculation.
First you must recognize that k1 is the largest K of the group and the pH of the solution will be due largely to the first dissociation. The second and third don't produce enough to concern us too much.
.......H3PO3 ==> H^+ + H2BO3^-
I......0.05M.....0.......0
C......-x........x.......x
E......0.05-x....x.......x
Substitute the E line into the Ka1 expression and solve for x = (H^+), then convert to pH.
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 and Explain any approximations or assumptions that you make in your calculation.
To determine the pH of a 0.050 M solution of boric acid (H3BO3), we need to consider the dissociation steps and their equilibrium constants.
1) First, let's calculate the concentration of each species at equilibrium after the first dissociation step:
[H3BO3] = 0.050 M (initial concentration)
[H^+] = x M (concentration of H^+ ions)
[H2BO3^-] = x M (concentration of H2BO3^- ions)
Using the equilibrium constant (K(a1)) for the first dissociation step:
K(a1) = [H^+][H2BO3^-] / [H3BO3]
7.3 x 10^-10 = x * x / (0.050 - x)
Since the initial concentration of H3BO3 is much greater than the x value, we can neglect x in the denominator:
7.3 x 10^-10 ≈ x^2 / 0.050
Simplifying the equation:
x^2 = 7.3 x 10^-10 * 0.050
x^2 = 3.65 x 10^-11
Taking the square root of both sides:
x = √(3.65 x 10^-11)
x ≈ 6.05 x 10^-6 M
2) Now, we can proceed to the second dissociation step. We assume that the concentration of [H2BO3^-] is the same as the x value obtained above since it's negligible compared to the initial concentration of H3BO3. Using the equilibrium constant (K(a2)) for the second dissociation step:
K(a2) = [H^+][HBO3^-2] / [H2BO3^-]
1.8 x 10^-13 = x * x / (6.05 x 10^-6)
Since the concentration of H2BO3^- is negligible compared to x, we can neglect it in the denominator:
1.8 x 10^-13 ≈ x^2 / (6.05 x 10^-6)
Simplifying the equation:
x^2 = 1.8 x 10^-13 * 6.05 x 10^-6
x^2 ≈ 1.089 x 10^-18
Taking the square root of both sides:
x ≈ √(1.089 x 10^-18)
x ≈ 1.043 x 10^-9 M
3) Finally, for the third dissociation step, we assume that the concentration of [HBO3^-] is the same as the x value obtained above since it's negligible compared to the initial concentration of H3BO3. Using the equilibrium constant (K(a3)) for the third dissociation step:
K(a3) = [H^+][BO3^-3] / [HBO3^-]
1.6 x 10^-14 = x * x / (1.043 x 10^-9)
Since the concentration of HBO3^- is negligible compared to x, we can neglect it in the denominator:
1.6 x 10^-14 ≈ x^2 / (1.043 x 10^-9)
Simplifying the equation:
x^2 = 1.6 x 10^-14 * (1.043 x 10^-9)
x^2 ≈ 1.671 x 10^-22
Taking the square root of both sides:
x ≈ √(1.671 x 10^-22)
x ≈ 1.293 x 10^-11 M
4) Finally, we can calculate the pH of the solution using the concentration of H^+ ions ([H^+]). Taking the negative logarithm of [H^+] will give us the pH:
pH = -log[H^+]
pH ≈ -log(1.293 x 10^-11)
pH ≈ 10.89
Therefore, the pH of a 0.050 M solution of boric acid is approximately 10.89.
Assumptions and approximations made in this calculation:
- We assumed that the initial concentration of H3BO3 is much greater than the concentrations of the dissociated species, so they can be neglected when calculating the equilibrium concentrations.
- We ignored the autoionization of water because the concentration of H^+ ions from water is much less than those produced from the dissociation of boric acid.
To calculate the pH of a 0.050 M solution of boric acid, we need to consider all three dissociation steps. Let's start by finding the concentration of H+ ions in each step.
Step 1:
H3BO3 ⇌ H+ + H2BO3-
Let's assume x is the concentration of H+ ions produced. Therefore, the concentration of H2BO3- will also be x. As a result, the concentration of H3BO3 decreases by x. We can set up an equation based on the equilibrium expression and the given Ka1 value:
Ka1 = [H+][H2BO3-] / [H3BO3]
7.3 x 10^-10 = x * x / (0.050 - x)
Since x is small compared to 0.050, we can make an approximation by assuming that (0.050 - x) ≈ 0.050. By doing this, we can simplify the equation as follows:
7.3 x 10^-10 ≈ x^2 / 0.050
Since Ka1 is very small, we can assume that the x value produced in this step is also negligible compared to 0.050. Therefore, we can neglect x in the denominator. Thus, our equation becomes:
7.3 x 10^-10 ≈ x^2 / 0.050
Solving this equation using quadratic formula or approximation methods, we find the concentration of H+ ions (x) in step 1.
Step 2:
H2BO3- ⇌ H+ + HBO3^-2
Assuming the concentration of H+ ions produced in this step is y, the concentration of HBO3^-2 will also be y. As a result, the concentration of H2BO3- decreases by y. Similarly, we can set up an equation based on the equilibrium expression and the given Ka2 value:
Ka2 = [H+][HBO3^-2] / [H2BO3-]
1.8 x 10^-13 = y * y / (0.050 - y)
Similarly, we can make the approximation (0.050 - y) ≈ 0.050 and solve the equation to find the concentration of H+ ions (y) in step 2.
Step 3:
HBO3 ⇌ H+ + BO3^-3
In this step, let's assume that z is the concentration of H+ ions produced, and the concentration of BO3^-3 will also be z. As a result, the concentration of HBO3 decreases by z. We set up an equation based on the equilibrium expression and the given Ka3 value:
Ka3 = [H+][BO3^-3] / [HBO3]
1.6 x 10^-14 = z * z / (0.050 - z)
Similarly, we can make the approximation (0.050 - z) ≈ 0.050 and solve the equation to find the concentration of H+ ions (z) in step 3.
Finally, we can calculate the total concentration of H+ ions by summing up the values obtained in steps 1, 2, and 3.
pH = -log[H+]
Please note that the approximations made, such as neglecting x, y, and z in the denominators, are based on the assumption that their values are much smaller compared to the initial concentration of boric acid (0.050 M). These approximations are valid under certain conditions, such as weak acid solutions and small ionization constants (Ka values). If the concentration of the acid is high or the Ka values are significant, these approximations may not be accurate, and a more precise method should be applied, such as solving the equilibrium equations using a numerical approach.