i need help solving this

Benzoic acid is a weak, monoprotic acid
(Ka = 6.3 × 10−5). Calculate the pH and the percent dissociation of each of the following solutions of benzoic acid. Then use Le Châtelier’s principle to explain the trend in percent dissociation of the acid as the solution becomes more dilute.
(a)1.0mol/L (b)0.10mol/L (c)0.01mol/L

I'll show you how to do a and I expect you can do the rest but I can help you through them if necessary. Call benzoic acid HBz.

..........HBz ==> H^+ + Bz^-
I.........1.0M....0......0
C..........-x.....x......x
E........1.0-x....x......x

Then 6.3E-5 = (x)(x)/(1.0-x)
You can do the math but I get x = approx 8E-3M = (H^+) and convert that to pH.
Then %dissoc = [(H^+)/(HBz)]*100 = approx [(8E-3)/1]*100 = about 0.8%

You can do b and c. (H^+) becomes less as (HBz) becomes smaller ( as you would expect) BUT % dissoc increases. Why? Because HBz + H2O ==> H3O^+ + Bz^-
As H2O is added (making the solution more dilute) Le Chatelier's Principle says the reaction will shift to the right and of course that means it is dissociated more in more dilute solutions.

Oh, acids and percent dissociation, huh? Well, let's see what we can do here.

To calculate the pH and percent dissociation, we can use the formula for percent dissociation (%D) which is equal to (concentration of dissociated acid / initial concentration of acid) × 100.

For (a) 1.0 mol/L:
Using the Ka value and knowing that [H+] = [A-], we can set up an equation:
Ka = [H+][A-] / [HA]
6.3 × 10^-5 = [H+][A-] / 1.0
[H+] = √(6.3 × 10^-5 × 1.0) = 0.0079 M (this is the concentration of H+)

Since we know that pH = -log[H+], we can calculate...
pH = -log(0.0079)
pH ≈ 2.1

For percent dissociation:
%D = ([H+]/[HA]) × 100
%D = (0.0079 / 1.0) × 100
%D ≈ 0.79%

Now, let's move on to the other solutions:

For (b) 0.10 mol/L:
[H+] ≈ 0.028 M
pH ≈ 1.55
%D ≈ 2.8%

For (c) 0.01 mol/L:
[H+] ≈ 0.089 M
pH ≈ 1.05
%D ≈ 8.9%

So, as the solution becomes more dilute, the percent dissociation of the acid increases. This can be explained using Le Châtelier’s principle. Dilution of a solution shifts the equilibrium to the right, favoring dissociation. It's like offering a clown a bigger clown wig - they just can't resist wearing it!

I hope that answered your question, or at least made you smile!

To solve this problem, we will use the formula for the percent dissociation of a weak acid:

% Dissociation = (concentration of dissociated acid / initial acid concentration) × 100

To calculate the pH of each solution, we will use the equation:

pH = -log[H+]

First, let's solve part (a):

(a) 1.0 mol/L:

Since 1.0 mol/L is a fairly concentrated solution, we can assume that the percent dissociation will be low. To find the percent dissociation, we need to calculate the concentration of dissociated acid. Since benzoic acid is a monoprotic acid, the concentration of the dissociated acid is equal to the concentration of H+ ions.

[H+] = √(Ka × C)

Where:
Ka = 6.3 × 10^(-5) (acid dissociation constant for benzoic acid)
C = 1.0 mol/L (concentration of acid)

[H+] = √(6.3 × 10^(-5) × 1.0) = 0.0079 mol/L

Now, let's calculate the percent dissociation:

% Dissociation = (0.0079 mol/L / 1.0 mol/L) × 100 = 0.79%

To calculate the pH, we will use the equation:

pH = -log[H+]

pH = -log(0.0079) ≈ 2.10

Therefore, the pH of the 1.0 mol/L solution of benzoic acid is approximately 2.10, and the percent dissociation is approximately 0.79%.

Now, let's move on to part (b):

(b) 0.10 mol/L:

As the solution becomes less concentrated, we can expect the percent dissociation to increase. Let's follow the same steps to calculate the percent dissociation and the pH:

[H+] = √(Ka × C) = √(6.3 × 10^(-5) × 0.10) ≈ 0.025 moles/L

% Dissociation = (0.025 moles/L / 0.10 moles/L) × 100 = 25%

pH = -log(0.025) ≈ 1.60

The pH of the 0.10 mol/L solution of benzoic acid is approximately 1.60, and the percent dissociation is approximately 25%.

Finally, let's solve part (c):

(c) 0.01 mol/L:

Since the concentration has decreased even more, we can expect a greater percent dissociation compared to before. Let's calculate:

[H+] = √(Ka × C) = √(6.3 × 10^(-5) × 0.01) ≈ 0.008 moles/L

% Dissociation = (0.008 moles/L / 0.01 moles/L) × 100 = 80%

pH = -log(0.008) ≈ 2.10

The pH of the 0.01 mol/L solution of benzoic acid is approximately 2.10, and the percent dissociation is approximately 80%.

Using Le Châtelier's principle, we can explain the trend in percent dissociation as the solution becomes more dilute. When the concentration decreases, the system shifts towards the dissociated form to restore equilibrium. Therefore, the percent dissociation increases as the solution becomes more dilute.

To solve this problem, we need to use the ionization constant (Ka) of benzoic acid and the formula for its percent dissociation.

The equation for the dissociation of benzoic acid is:
C6H5COOH ⇌ C6H5COO- + H+

First, let's calculate the pH of each solution using the Ka value of benzoic acid.

(a) 1.0 mol/L solution:

Step 1: Calculate the concentration of H+ ions (x) by assuming that x is the extent of dissociation.
C6H5COOH ⇌ C6H5COO- + H+
Initial concentration: 1.0 mol/L
Change in concentration: -x
Equilibrium concentration: (1.0 - x) mol/L

Since benzoic acid is a weak acid, we can neglect the change in concentration due to the dissociation.

Step 2: Use the Ka expression to set up an equation.
Ka = [C6H5COO-][H+] / [C6H5COOH]
Ka = x * x / (1.0 - x)

Step 3: Solve for x. Since Ka is relatively small, we can assume that the dissociation (x) is much smaller than 1.
Therefore, we can approximate 1.0 - x ≈ 1.0.

Ka = x * x / 1.0
6.3 × 10^(-5) = x^2
x ≈ 0.0079 (approximately)

Step 4: Calculate the pH using the concentration of H+ ions.
pH = -log[H+]
pH = -log(0.0079)
pH ≈ 2.10

So, the pH of the 1.0 mol/L solution of benzoic acid is approximately 2.10.

(b) 0.10 mol/L solution:

Follow the same steps as in part (a), but use 0.10 mol/L as the initial concentration.

Ka = x * x / (0.10 - x)
Solving for x, we find x ≈ 0.0079 (same as before).

pH = -log(0.0079)
pH ≈ 2.10

So, the pH of the 0.10 mol/L solution of benzoic acid is also approximately 2.10.

(c) 0.01 mol/L solution:

Again, follow the same steps as in part (a), but use 0.01 mol/L as the initial concentration.

Ka = x * x / (0.01 - x)
Solving for x, we find x = 0.0063.

pH = -log(0.0063)
pH ≈ 2.20

So, the pH of the 0.01 mol/L solution of benzoic acid is approximately 2.20.

Now, let's analyze the trend in the percent dissociation as the solution becomes more dilute using Le Châtelier's principle.

Le Châtelier's principle states that a system at equilibrium will shift in such a way as to counteract any imposed change and restore equilibrium.

In this case, as the solution becomes more dilute (lower concentration), the equilibrium reaction will shift to the left, favoring the undissociated form of benzoic acid (C6H5COOH).

As a result, the percent dissociation of benzoic acid will decrease as the solution becomes more dilute.

Therefore, the trend in percent dissociation of benzoic acid is that it decreases as the solution becomes more dilute.