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Chapter 14: Problem 196

An aqueous solution contains a mixture of 0.0500\(M \mathrm{HCOOH}\) \(\left(K_{\mathrm{a}}=1.77 \times 10^{-4}\right)\) and 0.150$M \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\left(K_{\mathrm{a}}=\right.\( \)1.34 \times 10^{-5} ) .\( Calculate the \)\mathrm{pH}$ of this solution. Because both acids are of comparable strength, the \(\mathrm{H}^{+}\) contribution from both acids must be considered.

Short Answer

Expert verified
The pH of the aqueous solution containing both 0.0500 M formic acid and 0.150 M propanoic acid is approximately 2.48. This was determined by calculating the H+ ion contributions from both acids, considering their comparable strength, and using their respective Ka values and initial concentrations.

Step by step solution

01

Identify the equilibrium expressions for both acids

Write the equilibrium expressions for both acids, using their respective dissociation constants (Ka): For formic acid (HCOOH): \( K_{a1} = \dfrac{[\mathrm{H}^{+}][\mathrm{HCOO}^{-}]}{[\mathrm{HCOOH}]} \) For propanoic acid (CH3CH2COOH): \( K_{a2} = \dfrac{[\mathrm{H}^{+}][\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COO}^{-}]}{[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COOH}]} \)
02

Set up ICE tables for both acids

For each acid, set up an Initial-Change-Equilibrium (ICE) table to find the equilibrium concentrations of H+ ions. We will denote the H+ ion contribution from formic acid as x, and the contribution from propanoic acid as y. Formic Acid: Initial: [H+]=x, [HCOO-]=0, [HCOOH]=0.0500-x, Change: -x, +x, -x, Equilibrium: 0, x, 0.0500-x Propanoic Acid: Initial: [H+]=y, [CH3CH2COO-]=0, [CH3CH2COOH]=0.150-y Change: -y, +y, -y, Equilibrium: 0, y, 0.150-y Now we will use these equilibrium concentrations in the Ka expressions to set up two equations to later solve for x and y.
03

Set up the equilibrium constant equations for both acids

Insert the equilibrium concentrations from the ICE tables into the Ka expressions for both acids: For formic acid: \( K_{a1} = 1.77 \times 10^{-4} = \dfrac{x \cdot x}{0.0500 - x} \) For propanoic acid: \( K_{a2} = 1.34 \times 10^{-5} = \dfrac{y \cdot y}{0.150 - y} \)
04

Solve for x and y, the H+ ion contributions from both acids

Solve each equilibrium constant equation for the H+ ion contributions (x and y) from both acids. For simpler problems, the quadratic formula could be used. In this case, as both acids have a comparable strength, we will solve them using numerical methods or a calculator, obtaining: x ≈ 2.85 × 10^(-3) M y ≈ 4.84 × 10^(-4) M
05

Calculate the total H+ ion concentration and pH

Now that we have the H+ ions' contributions from both acids, we can find the total H+ concentration in the solution by adding x and y: [H+] = x + y ≈ 2.85 × 10^(-3) M + 4.84 × 10^(-4) M ≈ 3.33 × 10^(-3) M Now, we can find the pH of the solution using the formula: \( \mathrm{pH} = - \log [\mathrm{H}^{+}] \) pH ≈ -log(3.33 × 10^(-3)) ≈ 2.48

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Most popular questions from this chapter

Consider a \(0.67-M\) solution of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)$ a. Which of the following are major species in the solution? i. \(C_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) ii.. \(\mathrm{OH}^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}+\) b. Calculate the \(\mathrm{pH}\) of this solution.

Calculate the concentration of all species present and the pH of a \(0.020-M\) HF solution.

Arsenic acid \(\left(\mathrm{H}_{3} \mathrm{AsO}_{4}\right)\) is a triprotic acid with \(K_{\mathrm{a}_{1}}=5.5 \times 10^{-3}\) $K_{\mathrm{a}_{2}}=1.7 \times 10^{-7},\( and \)K_{\mathrm{a}_{3}}=5.1 \times 10^{-12}$ . Calculate \(\left[\mathrm{H}^{+}\right]\) $\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right],\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right],\left[\mathrm{HAsO}_{4}^{2-}\right],$ and \(\left[\mathrm{AsO}_{4}^{3-]}\right]\) in a \(0.20-M\) arsenic acid solution.

Phosphoric acid is a common ingredient in traditional cola drinks. It is added to provide the drinks with a pleasant tart taste. Assuming that in cola drinks the concentration of phosphoric acid is \(0.007 M,\) calculate the \(\mathrm{pH}\) of this solution.

A certain acid, HA, has a vapor density of 5.11 \(\mathrm{g} / \mathrm{L}\) when in the gas phase at a temperature of \(25^{\circ} \mathrm{C}\) and a pressure of 1.00 atm. When 1.50 \(\mathrm{g}\) of this acid is dissolved in enough water to make 100.0 \(\mathrm{mL}\) of solution, the \(\mathrm{pH}\) is found to be $1.80 .\( Calculate \)K_{\mathrm{a}}$ for the acid.
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