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Chapter 14: Problem 123

Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.4 \times 10^{-5}\).

Short Answer

Expert verified
The pH of the 0.050 M Al(NO₃)₃ solution is approximately 3.28.

Step by step solution

01

Write the hydrolysis reaction

The hydrolysis reaction for Al(H₂O)₆³⁺ is as follows: \[ Al(H_2O)_6^{3+} + H_2O \rightleftharpoons H_3O^+ + Al(H_2O)_5(OH)^{2+} \]
02

Construct the ICE table

We will use an ICE (Initial, Change, Equilibrium) table to keep track of the changes in concentrations during the hydrolysis reaction: \[ \begin{array}{c|c|c|c} & [Al(H_2O)_6^{3+}] & [H_3O^+] & [Al(H_2O)_5(OH)^{2+}] \\ \hline I & 0.050 & 0 & 0 \\ C & -x & x & x \\ E & 0.050 - x & x & x \end{array} \]
03

Write the equilibrium expression and solve for x

Use the Ka value to solve for x. since Ka = 1.4 × 10⁻⁵. Therefore: \[ K_a = \frac{[H_3O^+][Al(H_2O)_5(OH)^{2+}]}{[Al(H_2O)_6^{3+}]} = \frac{x^2}{0.050 - x} \] Assuming x is significantly smaller than 0.050, we can simplify the equation as: \[ 1.4 \times 10^{-5} = \frac{x^2}{0.050} \]
04

Calculate the value of x

Solving for x, which represents the [H₃O⁺] concentration: \[ x = \sqrt{1.4 \times 10^{-5} \times 0.050} \approx 5.29 \times 10^{-4} \]
05

Calculate the pH

Use the pH formula: \[ pH = -\log[H_3O^+] = -\log(5.29 \times 10^{-4}) \]
06

Find the final pH value

Calculate the pH: \[ pH \approx 3.28 \] So, the pH of the 0.050 M Al(NO₃)₃ solution is approximately 3.28.

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Most popular questions from this chapter

A codeine-containing cough syrup lists codeine sulfate as a major ingredient instead of codeine. The Merck Index gives \(\mathrm{C}_{36} \mathrm{H}_{44} \mathrm{~N}_{2} \mathrm{O}_{10} \mathrm{~S}\) as the formula for codeine sulfate. Describe the composition of codeine sulfate. (See Exercise 143.) Why is codeine sulfate used instead of codeine?

Hemoglobin (abbreviated \(\mathrm{Hb}\) ) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is pH-dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}{ }^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}\), is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise 148.) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary?

An acid HX is \(25 \%\) dissociated in water. If the equilibrium concentration of \(\mathrm{HX}\) is \(0.30 \mathrm{M}\), calculate the \(K_{\mathrm{a}}\) value for \(\mathrm{HX}\).

Use the Lewis acid-base model to explain the following reaction. $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.050 \mathrm{M} \mathrm{NaCN}\)
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