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Chapter 16: Problem 81

What is the pH of \(0.25 M \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} ?\)

Short Answer

Expert verified
Answer: The pH of a 0.25 M Al(NO3)3 solution is 0.

Step by step solution

01

Write the chemical equation of Al3+ reacting with water

Al3+ will react with water to produce hydrogen ions (H+) and the aluminum hydroxide complex ion (Al(OH)4-). The balanced equation for this reaction is: Al3+ (aq) + 4H2O (l) <=> Al(OH)4- (aq) + 4H+ (aq)
02

Determine the equilibrium expression

Now, we will write the equilibrium expression for the reaction. The equilibrium constant (K) for this reaction can be calculated as: K = [Al(OH)4-][H+]^4 / [Al3+] Since the concentration of water is constant, it is not included in the equilibrium expression.
03

Find the concentration of H+ ions

The 0.25 M Al(NO3)3 solution means that the initial concentration of Al3+ ([Al3+]_initial) is 0.25 M. From the balanced equation, for every Al3+ that reacts, 4 H+ ions are produced. Therefore, we can set up the following expressions: [Al3+] = 0.25 - x [Al(OH)4-] = x [H+] = 4x Since K is significantly large, we can assume that all of the Al3+ ions have reacted. Therefore, the concentration of H+ ions [H+] will be: [H+] = 4 * (0.25) = 1
04

Calculate the pH

Once we have the concentration of H+ ions in the solution, we can then calculate the pH of the solution. The pH is defined as the negative logarithm of the hydrogen ion concentration: pH = -log[H+] For our 0.25 M Al(NO3)3 solution: pH = -log(1) = 0 The pH of the 0.25 M Al(NO3)3 solution is 0.

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Most popular questions from this chapter

What are the equilibrium concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{F}^{-}\) in a saturated solution of lead(II) fluoride at \(25^{\circ} \mathrm{C} ?\)

In which of the following solutions will \(\mathrm{CaF}_{2}\) be most soluble? (a) \(0.010 M \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} ;\) (b) \(0.01 M \mathrm{NaF}\) (c) \(0.001 M\) NaF; (d) \(0.10 M \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\)

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What is the difference between molar solubility and solubility product?

Why does a solution of a weak acid and its conjugate base control pH better than a solution of the weak acid alone?
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