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Chapter 13: Problem 75

A \(0.15-M\) solution of a weak acid is \(3.0 \%\) dissociated. Calculate \(K_{\mathrm{a}}\)

Short Answer

Expert verified
The equilibrium constant, \(K_a\), for the weak acid is approximately 1.39 × 10⁻⁴.

Step by step solution

01

Write the balanced equation for the dissociation of the weak acid

The general equation for the dissociation of a weak acid (HA) in water is: HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)
02

Calculate the equilibrium concentrations of all species from the given data

First, we need to calculate the equilibrium concentrations of all species involved in the reaction. We are given the initial concentration of the weak acid (0.15 M) and its percentage of dissociation (3.0 %). Let x = the concentration of H3O+ and A- at equilibrium: x = 0.15 M × 3.0 % = 0.15 M × 0.03 = 0.0045 M So, the equilibrium concentrations are: [H3O+] = [A-] = 0.0045 M [HA] = 0.15 M - 0.0045 M = 0.1455 M
03

Write the expression for the equilibrium constant, \(K_a\), and substitute equilibrium concentrations

The expression for \(K_a\) is: \(K_a\) = \(\frac{[H3O^+][A^-]}{[HA]}\) Now substitute the equilibrium concentrations we've found in Step 2: \(K_a\) = \(\frac{(0.0045)(0.0045)}{(0.1455)}\)
04

Calculate the value of \(K_a\)

To find the value of \(K_a\), simply multiply the numerator, divide by the denominator, and solve: \(K_a\) = \(\frac{(0.0045)(0.0045)}{(0.1455)}\) = \(\frac{0.00002025}{0.1455}\) = 1.39 × 10⁻⁴ The equilibrium constant, \(K_a\), for the weak acid is approximately 1.39 × 10⁻⁴.

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Most popular questions from this chapter

Arrange the following 0.10 \(M\) solutions in order from most acidic to most basic. See Appendix 5 for \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values. $$\mathrm{CaBr}_{2}, \quad \mathrm{KNO}_{2}, \quad \mathrm{HClO}_{4}, \quad \mathrm{HNO}_{2}, \quad \mathrm{HONH}_{3} \mathrm{ClO}_{4}$$

You have \(100.0 \mathrm{g}\) saccharin, a sugar substitute, and you want to prepare a \(\mathrm{pH}=5.75\) solution. What volume of solution can be prepared? For saccharin, \(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3}, \mathrm{p} K_{\mathrm{a}}=11.70\) \(\left(p K_{a}=-\log K_{a}\right)\)

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. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) iii. OH \(^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) b. Calculate the \(p\) H of this solution.

Calculate the pH of the following solutions: a. \(1.2 \mathrm{M} \mathrm{CaBr}_{2}\) b. \(0.84 M C_{6} H_{5} N H_{3} N O_{3}\left(K_{b} \text { for } C_{6} H_{5} N H_{2}=3.8 \times 10^{-10}\right)\) c. \(0.57 M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)\)

Which of the following represent conjugate acid-base pairs? For those pairs that are not conjugates, write the correct conjugate acid or base for each species in the pair. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) d. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)
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