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Chapter 16: Problem 17

What volume of \(6.15 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) is required to exactly neutralize 1.25 L of \(0.265 \mathrm{M} \mathrm{NH}_{3}(\text { aq }) ?\) $$\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}$$

Short Answer

Expert verified
53.9 mL of 6.15 M HCl solution is required to exactly neutralize 1.25 L of 0.265 M NH3 solution.

Step by step solution

01

Understanding the Reaction

First, we need to understand the reaction equation: \(\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\longrightarrow \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}\). Here, NH3 is acting as a base, and H3O+ (which can be released from the HCl in water) is functioning as an acid. In this case, the reaction involves a 1:1 stoichiometry, meaning one mole of NH3 reacts with one mole of H3O+.
02

Calculating the Moles of NH3

Now, we calculate the moles of NH3 in 1.25 L of 0.265 M NH3 solution. Based on the definition of Molarity, we find the moles by multiplying the volume by the molarity: \(1.25 \, \text{L} \times 0.265 \, \text{moles/L} = 0.33125 \, \text{moles}\).
03

Finding the Required Volume of HCl

Because of the 1:1 ratio in the reaction, 0.33125 moles of HCl is also required. To find the required volume of 6.15 M HCl, we divide the moles by the molarity: \(0.33125 \, \text{moles} ÷ 6.15 \, \text{moles/L} = 0.0539 \, \text{L} = 53.9 \, \text{mL}\).

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Most popular questions from this chapter

Which is the stronger acid of each of the following pairs of acids? Explain your reasoning. (a)\( HBr or HI;\) (b\() HOClO or HOBr; (c) I_SCCH_CH_COOH\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CCl}_{2} \mathrm{COOH}\).

Codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{N},\) is an opiate, has analgesic and antidiarrheal properties, and is widely used. In water, codeine is a weak base. A handbook gives \(\mathrm{p} K_{\mathrm{a}}=6.05\) for protonated codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{NH}^{+} .\) Write the reaction for \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{NH}^{+}\) and calculate \(\mathrm{p} K_{\mathrm{b}}\) for codeine.

The antimalarial drug quinine, \(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{O}_{2} \mathrm{N}_{2},\) is a diprotic base with a water solubility of \(1.00 \mathrm{g} / 1900 \mathrm{mL}\) of solution. (a) Write equations for the ionization equilibria corresponding to \(\mathrm{p} K_{\mathrm{b}_{1}}=6.0\) and \(\mathrm{p} K_{\mathrm{b}_{2}}=9.8\) (b) What is the \(\mathrm{pH}\) of saturated aqueous quinine?

You are asked to prepare a 100.0 mL sample of a solution with a pH of 5.50 by dissolving the appropriate amount of a solute in water with \(\mathrm{pH}=7.00 .\) Which of these solutes would you use, and in what quantity? Explain your choice. (a) \(15 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq}) ;\) (b) \(12 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) (c) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) ;\) (d) glacial (pure) acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\)

A 625 mL sample of an aqueous solution containing 0.275 mol propionic acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H},\) has \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.00239 \mathrm{M} .\) What is the value of \(K_{\mathrm{a}}\) for propionic acid? $$\begin{aligned} &\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2}\\\ &&K_{\mathrm{a}}=? \end{aligned}$$
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