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.