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Chapter 16: Problem 1

Use Le Châtelier's principle to explain how the common ion effect affects the \(\mathrm{pH}\) of a solution.

Short Answer

Expert verified
The Common ion effect influences the pH of a solution through Le Châtelier's principle. If a common ion is added to a solution it pushes the equilibrium either towards left or right, depending on whether it's an acid or base equilibrium, to minimize the change in concentration. This shifts affects the concentration of hydronium or hydroxide ions, subsequently altering the pH of the solution.

Step by step solution

01

Understand Le Châtelier's Principle

Le Châtelier's principle refers to the response of a chemical system at equilibrium when a change or stress is applied. The system will adjust to minimize the effect of the stress and restore equilibrium. The stress could be a change in concentration, pressure or temperature.
02

Understand the Common Ion Effect

The Common ion effect is a phenomenon that occurs when a weak acid or weak base solution is mixed with a strong acid or base that has an ion in common with the weak solution. Under this effect, the extent of ionization of the weak acid (or base) decreases, and hence the solution becomes less acidic (or basic). This is due to Le Châtelier's principle: the addition of a common ion pushes the equilibrium backward, resulting in fewer hydronium or hydroxide ions and subsequently affecting the pH.
03

Apply the Concepts

If, for example, a common ion is added to an acidic solution (an equilibrium system), Le Châtelier's principle states that the system will react to minimize the change. Therefore, the equilibrium will shift to the left, decreasing the concentration of H3O+ ions, thus increasing the pH.

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Most popular questions from this chapter

A \(10.0-\mathrm{mL}\) solution of \(0.300 \mathrm{M} \mathrm{NH}_{3}\) is titrated with a \(0.100 M \mathrm{HCl}\) solution. Calculate the \(\mathrm{pH}\) after the following additions of the \(\mathrm{HCl}\) solution: (a) \(0.0 \mathrm{~mL},\) (b) \(10.0 \mathrm{~mL}\) (c) \(20.0 \mathrm{~mL}\) (d) \(30.0 \mathrm{~mL}\) (e) \(40.0 \mathrm{~mL}\)

The molar solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}\) in a \(0.10 \mathrm{M} \mathrm{NaIO}_{3}\) solution is \(2.4 \times 10^{-11} \mathrm{~mol} / \mathrm{L} .\) What is \(K_{\mathrm{sp}}\) for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2} ?\)

In a titration experiment, \(12.5 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) neutralize \(50.0 \mathrm{~mL}\) of \(\mathrm{NaOH}\). What is the concentration of the \(\mathrm{NaOH}\) solution?

A student is asked to prepare a buffer solution at \(\mathrm{pH}=\) \(8.60,\) using one of the following weak acids: HA \(\left(K_{a}=2.7 \times 10^{-3}\right), \mathrm{HB}\left(K_{2}=4.4 \times 10^{-6}\right), \mathrm{HC}\left(K_{\mathrm{a}}=\right.\) \(2.6 \times 10^{-9}\) ). Which acid should she choose? Why?

For which of the following reactions is the equilibrium constant called a solubility product? $$ \begin{array}{l} \text { (a) } \mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{OH}^{-}(a q) \rightleftharpoons \\ \text { (b) } 3 \mathrm{Ca}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons \mathrm{Ca}_{3}(\mathrm{OH})_{4}^{2-}(a q) \end{array} $$ (c) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightleftharpoons\) $$ \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) $$ (d) \(\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q)\)
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