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Chapter 14: Problem 81

A solution of formic acid (HCOOH, \(K_{\mathrm{a}}=1.8 \times 10^{-4} )\) 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 0.02215 M.

Step by step solution

01

Write down the dissociation equation for formic acid

The dissociation equation for formic acid is: \( HCOOH \rightleftharpoons H^+ + HCOO^- \)
02

Write down the Ka expression from the equation

The Ka expression refers to the acid dissociation constant, which for formic acid, is given as \( 1.8 \times 10^{-4} \). This is derived from the concentrations of ions in the reaction: \( K_a = \frac{[H^+][HCOO^-]}{[HCOOH]} \)
03

Calculate the concentration of hydrogen ions using pH

We are given the pH of the solution (2.70). We can find the concentration of hydrogen ions, [H+], using the following formula: \( pH = -\log[H^+] \) We can rearrange this equation to solve for [H+]: \[ [H^+] = 10^{-pH} \] Now we plug in the pH value: \[ [H^+] = 10^{-2.7} = 1.995 \times 10^{-3} \]
04

Assume that formic acid's initial concentration and ion concentrations can be represented by variables

Let's assume the initial concentration of formic acid is "C". Since formic acid is a weak acid and it doesn't fully dissociate, we can represent the concentrations of the ions as follows: - [HCOOH] = C - x - [H+] = x - [HCOO-] = x
05

Substitute the variables into the Ka expression

Now we will substitute these values into the Ka expression: \( 1.8 \times 10^{-4} = \frac{x \cdot x}{C - x} \) We also know that the [H+] = x = 1.995 x 10^{-3} from Step 3, so we plug this value into the equation: \( 1.8 \times 10^{-4} = \frac{(1.995 \times 10^{-3})^2}{C - (1.995 \times 10^{-3})} \)
06

Solve the equation for C

Finally, we solve the equation for C (the initial concentration of formic acid): \[ C = \frac{(1.995 \times 10^{-3})^2}{1.8 \times 10^{-4}} + 1.995 \times 10^{-3} = 0.02215 \] Thus, the initial concentration of formic acid in the solution is 0.02215 M.

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Most popular questions from this chapter

Use the Lewis acid-base model to explain the following reaction. $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$

The pH of human blood is steady at a value of approximately 7.4 owing to the following equilibrium reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) $$ Acids formed during normal cellular respiration react with the \(\mathrm{HCO}_{3}^{-}\) to form carbonic acid, which is in equilibrium with \(\mathrm{CO}_{2}(a q)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) During vigorous exercise, a person's \(\mathrm{H}_{2} \mathrm{CO}_{3}\) blood levels were $26.3 \mathrm{mM},\( whereas his \)\mathrm{CO}_{2}\( levels were 1.63 \)\mathrm{mM}$ . On resting, the \(\mathrm{H}_{2} \mathrm{CO}_{3}\) levels declined to 24.9 \(\mathrm{m} M\) . What was the \(\mathrm{CO}_{2}\) blood level at rest?

Calculate the mass of sodium hydroxide that must be added to 1.00 \(\mathrm{L}\) of \(1.00-M \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) to double the pH of the solution (assume that the added NaOH does not change the volume of the solution).

Rank the following 0.10\(M\) solutions in order of increasing \(\mathrm{pH.}\) a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. $C_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}$ \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{HNO}_{3}\)

A typical aspirin tablet contains 325 mg acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right) .\) Calculate the \(\mathrm{pH}\) of a solution that is prepared by dissolving two aspirin tablets in enough water to make one \(\operatorname{cup}(237 \mathrm{mL})\) of solution. Assume the aspirin tablets are pure acetylsalicylic acid, \(K_{\mathrm{a}}=3.3 \times 10^{-4}\) .
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