To solve the problem, we need to address the two parts systematically:

Given Information:

• Concentration of the solution: $0.015 \, \text{M}$
• pH of the solution: $2.67$

i. Calculate the Hydronium Ion Concentration ($[H_3O^+]$):

We use the formula for pH: $\text{pH} = -\log[H_3O^+]$

Rearranging: $[H_3O^+] = 10^{-\text{pH}}$

Substitute $\text{pH} = 2.67$: $[H_3O^+] = 10^{-2.67} \approx 2.14 \times 10^{-3} \, \text{M}$

So, the hydronium ion concentration is: $[H_3O^+] = 2.14 \times 10^{-3} \, \text{M}$

ii. Calculate the Ionization Constant ($K_a$):

The ionization constant ($K_a$) for the acid can be found using the expression: [ K_a = \frac{[H_3O^+]^2}{[HA]_0 - [H_3O^+]} ]

Where:

• $[H_3O^+] = 2.14 \times 10^{-3} \, \text{M}$ (from part i),
• $[HA]_0 = 0.015 \, \text{M}$ (initial concentration of the acid).

Substitute the values: $K_a = \frac{(2.14 \times 10^{-3})^2}{0.015 - 2.14 \times 10^{-3}}$

First, calculate the numerator: $(2.14 \times 10^{-3})^2 = 4.58 \times 10^{-6}$

Next, calculate the denominator: $0.015 - 2.14 \times 10^{-3} = 0.01286$

Now, divide: $K_a = \frac{4.58 \times 10^{-6}}{0.01286} \approx 3.56 \times 10^{-4}$

So, the ionization constant is: $K_a \approx 3.56 \times 10^{-4}$

Final Answers:

1. Hydronium ion concentration: $2.14 \times 10^{-3} \, \text{M}$
2. Ionization constant ($K_a$): $3.56 \times 10^{-4}$

Would you like further clarifications or more details?

5 Related Questions:

1. How does temperature affect the ionization constant ($K_a$) of an acid?
2. What assumptions are made in the calculation of $K_a$?
3. Can the pH formula be used for very dilute solutions? Why or why not?
4. What is the relationship between $K_a$ and the strength of an acid?
5. How can you experimentally determine the pH of a solution?

Tip:

For weak acids, always check whether the degree of ionization is small enough to simplify $[HA]_0 - [H_3O^+] \approx [HA]_0$. This simplifies the math significantly for very weak acids!