Find the pH of solution (0.1M,100ml) HCOOH is Ka =10^2 and (0.1M,50ml)

To find the pH of the solutions you've mentioned, we need to consider the dissociation of formic acid (HCOOH) in water. Formic acid is a weak acid, and its dissociation can be represented by the following equilibrium reaction:

Understanding the Dissociation of Formic Acid

The dissociation of formic acid in water can be expressed as:

HCOOH ⇌ H⁺ + HCOO⁻

Given that the acid dissociation constant (Ka) for formic acid is 10⁻⁴, we can use this value to find the pH of the solutions.

Calculating pH for the First Solution (0.1 M, 100 mL)

For the first solution, we have 0.1 M formic acid in a volume of 100 mL. The concentration of HCOOH is 0.1 M, and we can set up an expression for the dissociation:

• Let x be the concentration of H⁺ ions produced at equilibrium.
• The initial concentration of HCOOH is 0.1 M, and at equilibrium, it will be (0.1 - x) M.
• The concentration of HCOO⁻ ions will also be x M.

Using the expression for Ka:

Ka = [H⁺][HCOO⁻] / [HCOOH]

Substituting the values we have:

10⁻⁴ = (x)(x) / (0.1 - x)

Assuming x is small compared to 0.1, we can simplify this to:

10⁻⁴ = x² / 0.1

Now, solving for x:

x² = 10⁻⁴ * 0.1 = 10⁻⁵

x = √(10⁻⁵) = 10^−2.5 ≈ 0.00316 M

Now, we can find the pH:

pH = -log[H⁺] = -log(0.00316) ≈ 2.50

Calculating pH for the Second Solution (0.1 M, 50 mL)

For the second solution, we have 0.1 M formic acid in a volume of 50 mL. The concentration remains the same, but the volume is halved. However, since the concentration is still 0.1 M, we can use the same approach as above.

Using the same Ka expression:

10⁻⁴ = (x)(x) / (0.1 - x)

Again, assuming x is small compared to 0.1, we simplify to:

10⁻⁴ = x² / 0.1

Solving for x gives us:

x² = 10⁻⁴ * 0.1 = 10⁻⁵

x = √(10⁻⁵) = 10^−2.5 ≈ 0.00316 M

Thus, the pH remains:

pH = -log(0.00316) ≈ 2.50

Final Thoughts

In both cases, the pH of the solutions is approximately 2.50. This demonstrates that the concentration of the acid does not affect the pH significantly when the acid is weak and the dissociation is minimal. If you have any more questions or need further clarification, feel free to ask!