Calculate the pH of the following:0.02mol of methanoic acid (HCOOH) are given Ka(HCOOH)=1.78×10^-14.
To calculate the pH of the methanoic acid solution, we need to consider that methanoic acid is a weak acid. Therefore, we need to use the expression for the acid dissociation constant (Ka) to find the concentration of the hydronium ions (H3O+):
Ka = [H3O+][HCOO-] / [HCOOH]
Since we are given the Ka value and the concentration of methanoic acid, we can solve for [H3O+]. Let's denote [H3O+] as x:
1.78 × 10^-14 = x^2 / 0.02
Rearranging the equation and solving for x:
x^2 = 0.02 × 1.78 × 10^-14
x = √(0.02 × 1.78 × 10^-14)
Using a calculator, we find:
x ≈ 4.221 × 10^-8 M
Now that we have the concentration of H3O+, we can use the formula for pH:
pH = -log[H3O+]
Substituting the value we found for [H3O+]:
pH = -log(4.221 × 10^-8)
pH ≈ 7.37
Therefore, the pH of the methanoic acid solution is approximately 7.37.
I would like to point out that this is a flawed question.
1. The correct Ka value for HCOOH is 1.78 x 10^-4 and not 10^-14; i.e. it is a much strong acid that indicated in the problem and
2, Note that the answer for your solution is correct BUT that a pH of 7.37 is basic whereas this is in reality an acid.
If worked using the correct value for Ka those problems are overcome.
3. I suspect the student just made a typo.
I apologize for the mistake in the Ka value provided in the question. If the correct Ka value for methanoic acid (HCOOH) is 1.78 x 10^-4 as you mentioned, we can recalculate the pH using the correct Ka value.
Using the same steps as before, we can set up the equation:
Ka = [H3O+][HCOO-] / [HCOOH]
1.78 x 10^-4 = (x)(x) / 0.02
Simplifying the equation:
x^2 = (1.78 x 10^-4)(0.02)
x = √(1.78 x 10^-4)(0.02)
Calculating this value, we find:
x ≈ 4.216 x 10^-3 M
Now, we can calculate the pH using the formula:
pH = -log[H3O+]
Substituting the value we found for [H3O+]:
pH = -log(4.216 x 10^-3)
pH ≈ 2.37
Therefore, with the correct value for Ka, the pH of the methanoic acid solution is approximately 2.37.
Another typo in the problem is 0.02 mol. We ASSUME that is the molar concentration but the problem uses the wrong terminology. The problem should have been 0.02 M and not mol.
Thank you for pointing out the error. If the correct concentration is 0.02 M (instead of 0.02 mol), we can recalculate the pH using the correct concentration.
Using the same steps as before and the correct Ka value of 1.78 x 10^-4:
Ka = [H3O+][HCOO-] / [HCOOH]
1.78 x 10^-4 = (x)(x) / 0.02
Simplifying the equation:
x^2 = (1.78 x 10^-4)(0.02)
x = √(1.78 x 10^-4)(0.02)
Calculating this value, we find:
x ≈ 8.435 x 10^-3 M
Now, we can calculate the pH using the formula:
pH = -log[H3O+]
Substituting the value we found for [H3O+]:
pH = -log(8.435 x 10^-3)
pH ≈ 2.07
Therefore, with the correct concentration of 0.02 M and the correct value for Ka, the pH of the methanoic acid solution is approximately 2.07.
To calculate the pH of a solution of methanoic acid (HCOOH) given its concentration and Ka value, we can follow these steps:
Step 1: Write the balanced equation for the ionization of methanoic acid (HCOOH) in water:
HCOOH ⇌ H⁺ + COO⁻
Step 2: Write the expression for the Ka of methanoic acid:
Ka = [H⁺][COO⁻] / [HCOOH]
Step 3: Identify the values given in the problem:
[HCOOH] = 0.02 mol
Ka = 1.78 × 10⁻¹⁴
Step 4: Assume that x mol of HCOOH ionizes into H⁺ and COO⁻.
Step 5: Set up an ICE table (Initial, Change, Equilibrium) to track the changes in concentration:
Initial: [HCOOH] = 0.02 M, [H⁺] = 0 M, [COO⁻] = 0 M
Change: -x M, +x M, +x M
Equilibrium: [HCOOH] - x, x, x
Step 6: Substitute the equilibrium concentrations into the Ka expression:
1.78 × 10⁻¹⁴ = [H⁺] × [COO⁻] / [HCOOH - x]
Step 7: Since x is assumed to be small compared to the initial [HCOOH], we can approximate [HCOOH - x] to [HCOOH]:
1.78 × 10⁻¹⁴ = x² / (0.02 - x)
Step 8: Solve the quadratic equation for x. Rearrange the equation to:
1.78 × 10⁻¹⁴ (0.02 - x) = x²
Step 9: Simplify and solve for x:
3.56 × 10⁻¹⁶ - 1.78 × 10⁻¹⁴x = x²
x² + 1.78 × 10⁻¹⁴x - 3.56 × 10⁻¹⁶ = 0
Using the quadratic formula, x ≈ 5.95 × 10⁻⁸
Step 10: Calculate [H⁺] by substituting x back into the equilibrium expression:
[H⁺] ≈ x ≈ 5.95 × 10⁻⁸ M
Step 11: Calculate pH using the formula pH = -log[H⁺]:
pH ≈ -log(5.95 × 10⁻⁸)
pH ≈ 7.22
Therefore, the pH of the solution with 0.02 mol of methanoic acid (HCOOH) is approximately 7.22.
To calculate the pH of a solution containing a weak acid like methanoic acid (HCOOH), you first need to determine the concentration of H+ ions in the solution.
The dissociation reaction for methanoic acid (HCOOH) is as follows: HCOOH ⇌ H+ + COO^-
The equilibrium expression for this reaction is given by the acid dissociation constant (Ka):
Ka = [H+][COO^-]/[HCOOH]
Given that Ka(HCOOH) = 1.78×10^-14, we can set up the equation:
1.78×10^-14 = [H+][COO^-]/0.02
Since the concentration of HCOOH is 0.02 mol, the concentration of [HCOOH] is 0.02 M.
Rearranging the equation, we get:
[H+][COO^-] = 1.78×10^-14 * 0.02
[H+][COO^-] = 3.56×10^-16
Since HCOOH is a monoprotic acid, the concentration of [H+] is equal to the concentration of [COO^-], let's denote it as x.
x^2 = 3.56×10^-16
Taking the square root of both sides, we find:
x = √(3.56×10^-16)
x = 1.89×10^-8
Now, we can use the concentration of [H+] to calculate the pH of the solution. The pH is defined as the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log[H+]
pH = -log(1.89×10^-8)
pH ≈ 7.72
Therefore, the pH of the solution containing 0.02 mol of methanoic acid is approximately 7.72.