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Chapter 14: Problem 77

A solution of formic acid \(\left(\mathrm{HCOOH}, K_{\mathrm{a}}=1.8 \times 10^{-4}\right)\) has a \(\mathrm{pH}\) of \(2.70 .\) Calculate the initial concentration of formic acid in this solution.

Short Answer

Expert verified
The initial concentration of formic acid in the solution is approximately 0.024 M.

Step by step solution

01

Write down the given information

We have the following given information: - pH of the solution: 2.70 - Ka of formic acid: 1.8 x 10^(-4)
02

Calculate the pKa

To find the pKa, use the formula pKa = -log10(Ka): pKa = -log10(1.8 x 10^(-4)) pKa ≈ 3.74
03

Use the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: pH = pKa + log([A-]/[HA]) We can rewrite the equation to solve for the ratio of conjugate base to formic acid concentrations: log([A-]/[HA]) = pH - pKa Plug in the values for pH and pKa from previous steps: log([A-]/[HA]) = 2.70 - 3.74 log([A-]/[HA]) = -1.04
04

Calculate the ratio of [A-] to [HA]

To find the ratio of [A-] to [HA], exponentiate both sides of the equation: [A-]/[HA] = 10^(-1.04) [A-]/[HA] ≈ 0.091
05

Set up the reaction stoichiometry

Here is the dissociation reaction of formic acid: HCOOH ↔ H+ + HCOO- Let the initial concentration of formic acid be 'x': Initially: [HCOOH] = x [H+] = 0 [HCOO-] = 0 Assuming that formic acid loses α moles of H+ ions in the dissociation, at equilibrium: [HCOOH] = x - α [H+] = α [HCOO-] = α Now, using the ratio of [A-] to [HA] from step 4, we know: α / (x - α) = 0.091
06

Solve for x, the initial concentration of formic acid

To find x, solve the equation: (α / (x - α)) = 0.091 Isolate the variable x: x - α = α / 0.091 x = α (1 + 1/0.091) Since the pH of the solution is given, we can write [H+] in terms of pH: [H+] = 10^(-pH) α = [H+] = 10^(-2.70) α ≈ 0.0020 Now, plug α into the equation for x: x = 0.0020 (1 + 1/0.091) x ≈ 0.0020 (1 + 11) x ≈ 0.0020 (12) x ≈ 0.024 Therefore, the initial concentration of formic acid in the solution is approximately 0.024 M.

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Most popular questions from this chapter

Using your results from Exercise 129, place the species in each of the following groups in order of increasing base strength. a. \(\mathrm{OH}^{-}, \mathrm{SH}^{-}, \mathrm{SeH}^{-}\) b. \(\mathrm{NH}_{3}, \mathrm{PH}_{3}\) c. \(\mathrm{NH}_{3}, \mathrm{HONH}_{2}\)

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