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Chapter 17: Problem 108

A sample of \(0.96 \mathrm{~L}\) of \(\mathrm{HCl}\) at \(372 \mathrm{mmHg}\) and \(22^{\circ} \mathrm{C}\) is bubbled into \(0.034 \mathrm{~L}\) of \(0.57 \mathrm{MH}_{3}\). What is the \(\mathrm{pH}\) of the resulting solution? Assume the volume of solution remains constant and that the \(\mathrm{HCl}\) is totally dissolved in the solution.

Short Answer

Expert verified
The pH of the resulting solution is 14.68.

Step by step solution

01

Determine the Moles of HCL

First, we have to calculate the number of moles of the HCl gas using the ideal gas law equation which is: \( PV = nRT \). Where P is pressure (in atm) so we convert \(372 \, mmHg \, to \, 0.49 \, atm\), V is volume(in Liters), n is the number of moles of gas. R is gas constant and its value is \(0.0821 \, L \cdot atm /mol \cdot K\), T is temperature in Kelvin (which we can get by adding 273 to the Celsius temperature). Thus, the number of moles of HCl = \( \frac{(0.49 \cdot 0.96)}{(0.0821 \cdot (22+273))} \)= 0.020 mol.
02

Determine the Moles of \( NH_{3} \)

Next, we have to get the number of moles of \( NH_{3} \) using the formula \(n = MV\), where M is molarity (in \(\frac{mol}{L}\)) and V is volume (in Liters). So, the moles of \( NH_{3} \)= \(0.57 \cdot 0.034\) = 0.019 mol.
03

Identify the limiting reactant and calculate \( [NH_{4}^{+}] \)

Based on the stoichiometry of the reaction \( NH_{3} + HCl \rightarrow NH_{4}^{+} + Cl^{-} \), both Hcl and \( NH_{3} \) react in a 1:1 ratio. Therefore, 0.019 moles of \( NH_{3} \) will react with 0.019 moles of HCL to form 0.019 moles of \( NH_{4}^{+} \) ion. HCL is the limiting reactant in this case.
04

Compute pOH

Since the solution is basic, we calculate the pOH using the formula \( pOH = -log[OH^{-}] \), by using \( pOH = -log \frac{[NH_{4}^{+}]}{V_{total}} \). The volume remains constant as mentioned in the problem. Therefore, \( pOH = -log \frac{0.019}{0.034+0.96} = -0.6759 \).
05

Calculate pH

To obtain the pH of the solution, we use the relation \( pH = 14 - pOH = 14 - (-0.6759) = 14.68 \).

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Most popular questions from this chapter

How does a common ion affect solubility? Use Le Châtelier's principle to explain the decrease in solubility of \(\mathrm{CaCO}_{3}\) in a \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution.

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with the carbonate and then "back-titrating" the remaining acid with \(\mathrm{NaOH}\). (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) were added to a \(0.1022-\mathrm{g}\) sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

A 5.00 -g quantity of a diprotic acid is dissolved in water and made up to exactly \(250 \mathrm{~mL}\). Calculate the molar mass of the acid if \(25.0 \mathrm{~mL}\) of this solution required \(11.1 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{KOH}\) for neutralization. Assume that both protons of the acid are titrated.

Explain how an acid-base indicator works in a titration.

A 1.0 -L saturated silver carbonate solution at \(5^{\circ} \mathrm{C}\) is treated with enough hydrochloric acid to decompose the compound. The carbon dioxide generated is collected in a \(19-\mathrm{mL}\) vial and exerts a pressure of \(114 \mathrm{mmHg}\) at \(25^{\circ} \mathrm{C}\). What is the \(K_{\mathrm{sp}}\) of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(5^{\circ} \mathrm{C} ?\)
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