How many moles of benzoic acid, a monoprotic acid with Ka=6.4×10^-5, must dissolve in 250mL of H2O to produce a solution with pH=2.04

To determine the number of moles of benzoic acid required to produce a solution with a pH of 2.04, we will need to use the concept of the acid dissociation constant (Ka) and the pH equation.

1. Recall the pH equation:
pH = -log[H+]

2. Rearrange the equation to solve for [H+]:
[H+] = 10^(-pH)

3. Since benzoic acid is a monoprotic acid, we can assume that [H+] is equal to the concentration of benzoic acid when it dissociates.

4. Write the dissociation reaction for benzoic acid:
C6H5COOH + H2O ⇌ C6H5COO- + H3O+

5. From the dissociation reaction, the benzoic acid concentration ([C6H5COOH]) can be assumed to be equal to [H3O+].

6. Calculate the concentration of H3O+ ions using the pH equation:
[H3O+] = 10^(-pH) = 10^(-2.04)

7. Convert the volume of water from milliliters to liters:
250 mL = 0.250 L

8. Calculate the number of moles of benzoic acid required using the concentration and volume:
Moles of benzoic acid = [H3O+] × volume of water
= 10^(-2.04) × 0.250 L

9. Calculate the result:
Moles of benzoic acid = 0.01016 moles

Therefore, approximately 0.01016 moles of benzoic acid must dissolve in 250 mL of water to produce a solution with a pH of 2.04.

To determine the number of moles of benzoic acid required, we can follow these steps:

Step 1: Find the concentration of H+ ions in the solution.
The pH of a solution is defined as the negative logarithm (base 10) of the concentration of H+ ions. Therefore, to find the concentration of H+ ions, we can use the formula:

[H+] = 10^(-pH)

In this case, the pH is given as 2.04. So,

[H+] = 10^(-2.04) = 0.006543 × 10^(-2) M

Step 2: Use the Ka expression to find the concentration of the benzoic acid (C6H5COOH) and the concentration of the conjugate base (C6H5COO-) in the solution.

Ka is the acid dissociation constant and is given as 6.4×10^(-5).

For a monoprotic acid, such as benzoic acid, the dissociation reaction is:

C6H5COOH ⇌ C6H5COO- + H+

The initial concentration of benzoic acid is x (in moles).

At equilibrium, the concentration of benzoic acid is (x - [H+]) and the concentration of C6H5COO- is [H+].

Therefore, we can write the expression for Ka:

Ka = ([C6H5COO-] * [H+]) / [C6H5COOH]

Since the concentration of C6H5COO- is equal to the concentration of H+, we can rewrite the expression as:

Ka = ([H+]^2) / [C6H5COOH]

Substituting the values we have:

6.4×10^(-5) = (0.006543 × 10^(-2))^2 / [C6H5COOH]

Step 3: Solve for [C6H5COOH].

Let's solve for [C6H5COOH]:

6.4×10^(-5) = (0.006543 × 10^(-2))^2 / [C6H5COOH]

Simplifying:

[C6H5COOH] = (0.006543 × 10^(-2))^2 / 6.4×10^(-5)

[C6H5COOH] = 0.006543^2 × 10^(-4) / 6.4×10^(-5)

[C6H5COOH] = 8.428 × 10^(-8) M

Step 4: Calculate the moles of C6H5COOH required.

We know that the volume of the solution is 250 mL, which can be converted to liters:

Volume (V) = 250 mL = 0.250 L

To find the number of moles of C6H5COOH, we can use the formula:

Moles = Concentration * Volume

Moles = 8.428 × 10^(-8) M * 0.250 L

Moles = 2.107 × 10^(-8) mol

Therefore, approximately 2.107 × 10^(-8) moles of benzoic acid must dissolve in 250 mL of H2O to produce a solution with a pH of 2.04.

Call benzoic acid HA, then

.....HA --> H^+ + A^-
E.....x.....

pH = -log(2.04)
Ka = (H^+)(A^-)/HA)
You know H^+. A^- is the same. You know Ka. Solve for x = HA. I get approximately 1.6 mols/L. For 250 mL you would need 1/4 that amount.