Just two more to check, thanks!

1) The pH of a 1.25 M solution of unknown monoprotic acid is 5.43. Determine the value of Ka for this acid. HX <--> H+ + X-. (I obtain 1.0 x 10^-11)

2) Calculate the pH of a solution which is 0.73 M benzoic acid. Ka = 6.5 x 10-5, HC7H5O2 <--> H+ + C7H5O2- (Another icebox problem, I obtain 2.2)

I agree with #2. For #1 I obtained 1.1E-11

I expect the difference in #1 is that I didn't read anything on the calculator but kept everything in until the final answer.

Sure, I can help you with both questions.

1) To determine the value of Ka for the unknown monoprotic acid, we need to use the equation: Ka = ( [H+][X-] ) / [HX]

Given that the pH of the solution is 5.43, we can calculate the concentration of H+ using the formula: [H+] = 10^(-pH).

[H+] = 10^(-5.43) = 2.74 x 10^(-6) M

Since the acid is monoprotic, the concentration of X- is equal to the concentration of H+, which is 2.74 x 10^(-6) M.

We also know that the initial concentration of HX is 1.25 M.

Using the given values, we can calculate Ka:

Ka = ( [H+][X-] ) / [HX] = (2.74 x 10^(-6) M) * (2.74 x 10^(-6) M) / (1.25 M)

Ka = 5.984 x 10^(-12) ≈ 6.0 x 10^(-12)

Therefore, the value of Ka for the unknown monoprotic acid is approximately 6.0 x 10^(-12).

2) To calculate the pH of the solution with a concentration of 0.73 M benzoic acid, we can use the Ka value given.

Given Ka = 6.5 x 10^(-5), the equation for benzoic acid dissociation is: HC7H5O2 <--> H+ + C7H5O2-

Let's assume x moles of benzoic acid dissociate. As a result, we will have x moles of H+ and x moles of C7H5O2-.

The initial concentration of the benzoic acid, [HC7H5O2], is 0.73 M. At equilibrium, the concentration of H+ will also be x M.

Using the equilibrium equation, Ka = ( [H+][C7H5O2-] ) / [HC7H5O2], we can substitute the given values:

6.5 x 10^(-5) = ( x * x ) / (0.73 - x)

Since x is much smaller than 0.73, we can assume that (0.73 - x) is approximately 0.73.

6.5 x 10^(-5) = x^2 / 0.73

Rearranging the equation, we get:

x^2 = (6.5 x 10^(-5)) * 0.73

x^2 = 4.745 x 10^(-5)

Taking the square root of both sides, we find:

x = 2.177 x 10^(-3) M

The concentration of H+ is equal to x, so the pH can be calculated as:

pH = -log[H+] = -log(2.177 x 10^(-3)) ≈ 2.26

Therefore, the pH of the 0.73 M benzoic acid solution is approximately 2.26.

Sure, I can help you with these problems and explain the steps to get the answers.

1) To determine the value of Ka for the unknown monoprotic acid, you can use the equation for the acid dissociation constant (Ka):
Ka = [H+][X-] / [HX]
You are given the pH of the solution, which is 5.43. To find the concentration of H+ ions, you can use the equation:
[H+] = 10^(-pH)

In this case, [H+] = 10^(-5.43)

Since it is a monoprotic acid, the concentration of X- ions is the same as the concentration of H+ ions. So, [X-] = 10^(-5.43).

The concentration of the acid (HX) can be calculated by using the formula:
[H+] + [X-] = [HX]
Substituting the values, you get:
10^(-5.43) + 10^(-5.43) = [HX]

Finally, by substituting the concentrations of H+, X-, and HX into the equation for Ka, you can calculate the value of Ka.

2) To calculate the pH of a solution containing benzoic acid, you can use the Henderson-Hasselbalch equation, which relates the pH of a weak acid solution to its pKa and the ratio of the conjugate base to the acid form:
pH = pKa + log([C7H5O2-]/[HC7H5O2])

In this case, the pKa of benzoic acid is given as 6.5, and the concentration of HC7H5O2 (benzoic acid) is 0.73 M.

Before applying the Henderson-Hasselbalch equation, you need to calculate the ratio [C7H5O2-]/[HC7H5O2]. The concentration of benzoic acid remains the same, so the ratio will just be [C7H5O2-]/0.73.

By substituting the given values into the Henderson-Hasselbalch equation, you can solve for the pH of the solution.

I hope this helps you understand how to solve these problems! Let me know if you have any further questions.