Chapter 14: Problem 3
Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of each solution added change the effectiveness of the buffer?
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Chapter 14: Problem 3
Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of each solution added change the effectiveness of the buffer?
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An aqueous solution contains dissolved \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} .\) The concentration of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) is \(0.50 M\) and \(\mathrm{pH}\) is 4.20 a. Calculate the concentration of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) in this buffer solution. b. Calculate the \(\mathrm{pH}\) after \(4.0 \mathrm{g} \mathrm{NaOH}(s)\) is added to \(1.0 \mathrm{L}\) of this solution. (Neglect any volume change.)
A friend asks the following: "Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like NaOH is added, the HA reacts with the OH - to form A Thus the amount of acid (HA) is decreased, and the amount of base \(\left(\mathrm{A}^{-}\right)\) is increased. Analogously, adding HCl to the buffered solution forms more of the acid (HA) by reacting with the base \(\left(\mathrm{A}^{-}\right)\). Thus how can we claim that a buffered solution resists changes in the pH of the solution?" How would you explain buffering to this friend?
Calculate the \(\mathrm{pH}\) of each of the following buffered solutions. a. \(0.50 M C_{2} H_{5} N H_{2} / 0.25 M C_{2} H_{5} N H_{3} C l\) b. \(0.25 M C_{2} H_{5} N H_{2} / 0.50 M C_{2} H_{5} N H_{3} C l\) c. \(0.50 M C_{2} H_{5} N H_{2} / 0.50 M C_{2} H_{5} N H_{3} C l\)
In the titration of \(50.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) \(\left(K_{\mathrm{b}}=4.4 \times 10^{-4}\right),\) with \(0.50 M\) HCl, calculate the pH under the following conditions. a. after \(50.0 \mathrm{mL}\) of \(0.50 \mathrm{M}\) HCl has been added b. at the stoichiometric point
Repeat the procedure in Exercise \(61,\) but for the titration of \(25.0 \mathrm{mL}\) of 0.100 \(\mathrm{M}\) \(\mathrm{HNO}_{3}\) with 0.100 \(\mathrm{M}\) \(\mathrm{NaOH}\).
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