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Chapter 14: Problem 39

Calculate the \(\mathrm{pH}\) after 0.010 mole of gaseous \(\mathrm{HCl}\) is added to \(250.0 \mathrm{mL}\) of each of the following buffered solutions. a. \(0.050 M \mathrm{NH}_{3} / 0.15 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) b. \(0.50 M \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?

Short Answer

Expert verified
The pH of both buffered solutions after adding 0.010 mol of HCl is approximately 8.98. The two original buffered solutions have the same pH but differ in capacity. The 0.50 M NH₃ / 1.50 M NH₄Cl buffered solution has a greater capacity compared to the 0.050 M NH₃ / 0.15 M NH₄Cl buffered solution, which makes it more efficient and resistant to pH changes when more acid or base is added to the system.

Step by step solution

01

Calculate the initial moles of NH₃ and NH₄Cl

In a 250 mL solution: Moles of NH₃ = Molarity × Volume = 0.050 M × 0.250 L = 0.0125 mol Moles of NH₄Cl = Molarity × Volume = 0.15 M × 0.250 L = 0.0375 mol
02

Reaction between HCl and NH₃

When 0.010 mol of gaseous HCl is added, it will react with NH₃ to form NH₄⁺ ions. As the amount of HCl is less than NH₃, 0.010 mol of NH₃ will react with 0.010 mol of HCl to form 0.010 mol of NH₄⁺. Moles of NH₃ after reaction = 0.0125 - 0.010 = 0.0025 mol Moles of NH₄⁺ after reaction = 0.0375 + 0.010 = 0.0475 mol
03

Use the Henderson-Hasselbalch equation to find pH

The Henderson-Hasselbalch equation is: pH = pK_a + log ([NH₃]/[NH₄⁺]) The pK_a of NH₄⁺ is 9.25. Hence, the pH = 9.25 + log (0.0025/0.0475) = 9.25 + log(0.05263) ≈ 8.98 #b. 0.50 M NH₃ / 1.50 M NH₄Cl buffered solution#
04

Calculate the initial moles of NH₃ and NH₄Cl

In a 250 mL solution: Moles of NH₃ = Molarity × Volume = 0.50 M × 0.250 L = 0.125 mol Moles of NH₄Cl = Molarity × Volume = 1.50 M × 0.250 L = 0.375 mol
05

Reaction between HCl and NH₃

When 0.010 mol of gaseous HCl is added, it will react with NH₃ to form NH₄⁺ ions. As the amount of HCl is less than NH₃, 0.010 mol of NH₃ will react with 0.010 mol of HCl to form 0.010 mol of NH₄⁺. Moles of NH₃ after reaction = 0.125 - 0.010 = 0.115 mol Moles of NH₄⁺ after reaction = 0.375 + 0.010 = 0.385 mol
06

Use the Henderson-Hasselbalch equation to find pH

The Henderson-Hasselbalch equation is: pH = pK_a + log ([NH₃]/[NH₄⁺]) Hence, the pH = 9.25 + log (0.115/0.385) = 9.25 + log(0.2987) ≈ 8.98
07

Comparison of pH and capacity between the two buffered solutions

The two original buffered solutions have the same pH values (8.98), but their capacities are different. A buffer with greater capacity contains more moles of NH₃ and NH₄⁺ ions, allowing it to maintain the pH even when more acid or base is added. The advantage of having a buffer with greater capacity is to maintain the pH under a broader range of conditions, making it more efficient and resistant to pH changes when more acid or base is added to the system. In this case, the 0.50 M NH₃ / 1.50 M NH₄Cl buffered solution has a greater capacity compared to the 0.050 M NH₃ / 0.15 M NH₄Cl buffered solution.

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Most popular questions from this chapter

A buffer is made using \(45.0 \mathrm{mL}\) of \(0.750 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=\right.\) \(1.3 \times 10^{-5}\) ) and \(55.0 \mathrm{mL}\) of \(0.700 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} .\) What volume of 0.10 \(M\) NaOH must be added to change the pH of the original buffer solution by \(2.5 \% ?\)

What quantity (moles) of HCl(g) must be added to \(1.0 \mathrm{L}\) of \(2.0 \mathrm{M}\) NaOH to achieve a pH of 0.00? (Neglect any volume changes.)

Consider the titration of \(80.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) by \(0.400 M\) HCl. Calculate the \(p H\) of the resulting solution after the following volumes of HCl have been added. a. \(0.0 \mathrm{mL}\) b. \(20.0 \mathrm{mL}\) c. \(30.0 \mathrm{mL}\) d. \(40.0 \mathrm{mL}\) e. \(80.0 \mathrm{mL}\)

A certain buffer is made by dissolving \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

Consider the titration of \(40.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{HClO}_{4}\) by \(0.100 M\) KOH. Calculate the \(p\) H of the resulting solution after the following volumes of KOH have been added. a. \(0.0 \mathrm{mL}\) b. \(10.0 \mathrm{mL}\) c. \(40.0 \mathrm{mL}\) d. \(80.0 \mathrm{mL}\) e. \(100.0 \mathrm{mL}\)
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