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Chapter 13: Problem 190

An aqueous solution contains a mixture of 0.0500 \(M\) HCOOH \(\left(K_{\mathrm{a}}=1.77 \times 10^{-4}\right)\) and \(0.150 M \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\left(K_{\mathrm{a}}=1.34 \times\right.\) \(10^{-5}\) ). Calculate the \(p\) H 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 0.0500 $M$ HCOOH and 0.150 $M$ CH3CH2COOH is approximately 2.44.

Step by step solution

01

Write the ionization reactions for both acids

The ionization reactions for each acid are as follows: For HCOOH: \(HCOOH \rightleftharpoons H^+ + HCOO^-\) For CH3CH2COOH: \(CH_{3}CH_{2}COOH \rightleftharpoons H^+ + CH_{3}CH_{2}COO^-\)
02

Set up the initial and equilibrium concentrations for both reactions

Let x = [H+] contributed by HCOOH Let y = [H+] contributed by CH3CH2COOH | HCOOH | H+ | HCOO- | Initial| 0.0500| 0 | 0 | Change | - x | +x | + x | Final |0.0500-x| x | x | | CH3CH2COOH | H+ | CH3CH2COO- | Initial| 0.150 | 0 | 0 | Change | -y | +y | +y | Final | 0.150-y | y | y |
03

Write the Ka expressions for both acids

For HCOOH: \(K_{a1} = 1.77 \times 10^{-4} = \frac{x^2}{0.0500 - x}\) For CH3CH2COOH: \(K_{a2} = 1.34 \times 10^{-5} = \frac{y^2}{0.150 - y}\)
04

Use the quadratic formula to find the equilibrium [H+] concentration

Solve for x and y using the quadratic formulas, such as the quadratic equation in terms of x: \(x^2 + 1.77 \times 10^{-4}x - 8.85 \times 10^{-6} = 0\) Solve for x, resulting in: \(x \approx 3.30 \times 10^{-3}\) Similarly, solve for y, resulting in: \(y \approx 3.56 \times 10^{-4}\) Sum the contributions of [H+] from both acids: \(x + y = 3.3 \times 10^{-3} + 3.56 \times 10^{-4} \approx 3.66 \times 10^{-3}\)
05

Calculate the pH

Now, we can find the pH using the final [H+] concentration: \(pH = - \log{([H^+])} = - \log{(3.66 \times 10^{-3})} \approx 2.44\) Therefore, the pH of the solution is approximately 2.44.

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Most popular questions from this chapter

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a derivative of morphine that is used as an analgesic, narcotic, or antitussive. It was once commonly used in cough syrups but is now available only by prescription because of its addictive properties. If the \(\mathrm{pH}\) of a \(1.7 \times 10^{-3}-M\) solution of codeine is \(9.59,\) calculate \(K_{\mathrm{b}}\)

A 10.0 -mL sample of an HCl solution has a pH of \(2.000 .\) What volume of water must be added to change the pH to 4.000?

Aluminum hydroxide is an amphoteric substance. It can act as either a Brönsted-Lowry base or a Lewis acid. Write a reaction showing Al(OH) \(_{3}\) acting as a base toward \(\mathrm{H}^{+}\) and as an acid toward OH \(^{-}\)

Consider \(0.25 \mathrm{M}\) solutions of the following salts: \(\mathrm{NaCl}\), RbOCI, KI, Ba(CIO_{ } _ { 2 } \text { , and } \mathrm { NH } _ { 4 } \mathrm { NO } _ { 3 } \text { . For each salt, indicate } whether the solution is acidic, basic, or neutral.

A typical sample of vinegar has a pH of \(3.0 .\) Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{a}=1.8 \times\right.\) \(10^{-5}\) ), calculate the concentration of acetic acid in vinegar.
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