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Chapter 15: Problem 101

What concentration of \(\mathrm{NH}_{4} \mathrm{Cl}\) is necessary to buffer a \(0.52-M\) \(\mathrm{NH}_{3}\) solution at \(\mathrm{pH}=9.00 ?\left(K_{\mathrm{b}}\right.\) for \(\left.\mathrm{NH}_{3}=1.8 \times 10^{-5} .\right)\)

Short Answer

Expert verified
To buffer a 0.52 M NH3 solution at pH 9.00, we first calculate pOH (5.00) and pKa from the given Kb. We then use the Henderson-Hasselbalch equation to find the concentration of NH4+ ions: [NH4+] / [NH3] = 10^(pOH - pKa). Finally, we determine the concentration of NH4Cl as [NH4+] = 0.52 × 10^(5.00 - pKa).

Step by step solution

01

Determine the pOH

Since we are given the pH of the solution, we can easily determine the pOH value by using the relationship between pH and pOH: pOH = 14 - pH From the given pH value, the pOH can be calculated as: pOH = 14 - 9.00 = 5.00
02

Calculate the pKa of the solution

We are given the Kb of NH3, now we'll find the Ka or the acidity constant as we will use pKa in the Henderson-Hasselbalch equation. The relationship between Ka, Kb, and Kw (ion product of water) can be expressed as: Ka × Kb = Kw We know the value of Kb (1.8 x 10^-5) and Kw (1 x 10^-14). Now, we can calculate the Ka value for the given system: Ka = Kw / Kb = (1 x 10^-14) / (1.8 x 10^-5) From this, we can now calculate the pKa value: pKa = -log(Ka)
03

Use the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is written as: pOH = pKa + log ([NH4+] / [NH3]) We have found the pOH and pKa values. We are given the concentration of NH3 (0.52 M). Now we can rearrange the equation to find the concentration of NH4+ ions or the concentration of NH4Cl: [NH4+] / [NH3] = 10^(pOH - pKa)
04

Calculate the concentration of NH4Cl

Now we will use the values of pOH, pKa, and NH3 to find the concentration of NH4+ ions: [NH4+] / 0.52 = 10^(5.00 - pKa) To find the concentration of NH4Cl, we just need to find [NH4+], and we can do this by solving for [NH4+]: [NH4+] = 0.52 × 10^(5.00 - pKa) Finally, calculate the concentration of NH4Cl needed to buffer the given NH3 solution at pH 9.00.

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Most popular questions from this chapter

When a diprotic acid, \(\mathrm{H}_{2} \mathrm{~A}\), is titrated with \(\mathrm{NaOH}\), the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added, what volume of \(\mathrm{NaOH}\) added corresponds to the second equivalence point? b. For the following volumes of \(\mathrm{NaOH}\) added, list the major species present after the \(\mathrm{OH}^{-}\) reacts completely. i. \(0 \mathrm{~mL}\) NaOH added ii. between 0 and \(100.0 \mathrm{~mL}\) NaOH added iii. \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added iv. between \(100.0\) and \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added v. \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added vi. after \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added c. If the \(\mathrm{pH}\) at \(50.0 \mathrm{~mL} \mathrm{NaOH}\) added is \(4.0\) and the \(\mathrm{pH}\) at \(150.0 \mathrm{~mL} \mathrm{NaOH}\) added is \(8.0\), determine the values \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for the diprotic acid.

Calculate the pH of each of the following solutions. a. \(0.100 \mathrm{M} \mathrm{HONH}_{2}\left(K_{\mathrm{b}}=1.1 \times 10^{-8}\right)\) b. \(0.100 \mathrm{M} \mathrm{HONH}_{3} \mathrm{Cl}\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 \mathrm{M} \mathrm{HONH}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{HONH}_{3} \mathrm{Cl}\)

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 \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HCl}\) have been added. a. \(0.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) c. \(30.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) e. \(80.0 \mathrm{~mL}\)

Consider the titration of \(40.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HClO}_{4}\) by \(0.100 M \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) 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. \(100.0 \mathrm{~mL}\)

Calculate the \(\mathrm{pH}\) of a solution that is \(0.40 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) and \(0.80 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3} .\) In order for this buffer to have \(\mathrm{pH}=\) \(\mathrm{p} K_{\mathrm{a}}\), would you add \(\mathrm{HCl}\) or \(\mathrm{NaOH}\) ? What quantity (moles) of which reagent would you add to \(1.0 \mathrm{~L}\) of the original buffer so that the resulting solution has \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) ?
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