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Chapter 15: Problem 113

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

Short Answer

Expert verified
The concentration of $\mathrm{NH}_{4} \mathrm{Cl}$ needed to buffer a $0.52 \mathrm{M}$ $\mathrm{NH}_{3}$ solution at pH 9.00 is \(9.36 \times 10^{-4}\, M\).

Step by step solution

01

Find the pOH and the OH⁻ concentration

Since we are given the pH, we will first find the pOH using the relationship: \[pH + pOH = pK_w = 14\] where \(pK_w\) is the ionic product of water. Now we can calculate the pOH: \[pOH = 14 - pH = 14 - 9 = 5\] Now that we have the pOH, we can calculate the \(OH^-\) concentration using the formula: \[OH^- = 10^{-pOH}\] \[OH^- = 10^{-5}\]
02

Set up the Kb expression

For NH\(_3\) the reaction equation is: \[NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\] Given the base dissociation constant \(K_b\) for NH\(_3\): \[K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\] We need to find the concentration of \(NH_4^+\).
03

Use the Kb expression to find the concentration of NH4⁺

Using the equation from step 2, we can substitute the values of \(K_b\), \([OH^-]\), and \([NH_3]\). \[ K_b = 1.8 \times 10^{-5} = \frac{[NH_4^+](10^{-5})}{0.52}\] Next, we need to solve for the \([NH_4^+]\): \[ [NH_4^+] = 1.8 \times 10^{-5} \times \frac{0.52}{10^{-5}}\] \[ [NH_4^+] = 9.36 \times 10^{-4} M\]
04

Determine the concentration of NH4Cl needed

The concentration of NH\(_4\)Cl is equal to the concentration of the NH\(_4^+\) ion released in the solution, so the concentration of NH\(_4\)Cl needed to buffer the 0.52 M NH\(_3\) solution at pH 9.00 is: \[ [NH_4Cl] = [NH_4^+] = 9.36 \times 10^{-4} M\]

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Most popular questions from this chapter

Amino acids are the building blocks for all proteins in our bodies. A structure for the amino acid alanine is All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has \(K_{\mathrm{a}}=4.5 \times 10^{-3}\) and the amino group has \(K_{\mathrm{b}}=7.4 \times 10^{-5} .\) Because of the two groups with acidic or basic properties, three different charged ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with \(\left[\mathrm{H}^{+}\right]=1.0 M ?\) In a solution with \(\left[\mathrm{OH}^{-}\right]=1.0 \mathrm{M} ?\)

A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.75 \(\mathrm{mL}\) of the weak acid solution has been added to 50.0 \(\mathrm{mL}\) of the 0.100 \(\mathrm{M}\) NaOH solution, the \(\mathrm{pH}\) of the resulting solution is 10.50 . Calculate the original concentration of the solution of weak acid.

Consider the titration of 150.0 \(\mathrm{mL}\) of 0.100 $\mathrm{M} \mathrm{HI}\( by 0.250 \)\mathrm{M}\( \)\mathrm{NaOH}$ . a. Calculate the pH after 20.0 \(\mathrm{mL}\) of NaOH has been added. b. What volume of NaOH must be added so that the \(\mathrm{pH}=7.00 ?\)

Consider the titration of 40.0 \(\mathrm{mL}\) of 0.200 $\mathrm{M} \mathrm{HClO}_{4}$ by 0.100 \(\mathrm{M}\) KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 80.0 \mathrm{mL}} \\ {\text { b. } 10.0 \mathrm{mL}} & {\text { e. } 100.0 \mathrm{mL}} \\ {\text { c. } 40.0 \mathrm{mL}}\end{array} $$

Repeat the procedure in Exercise \(67,\) 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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