# 8.5 Hydrolysis of salt solutions  (Page 5/5)

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$\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{O}\text{H}\right)}^{2+}\left(aq\right)$
$\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{2+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{4}\left({\text{OH}\right)}_{2}{}^{\text{+}}\left(aq\right)$
$\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{4}\left({\text{OH}\right)}_{2}{}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{3}\left({\text{OH}\right)}_{3}\left(aq\right)$

Note that some of these aluminum species are exhibiting amphiprotic behavior, since they are acting as acids when they appear on the left side of the equilibrium expressions and as bases when they appear on the right side.

However, the ionization of a cation carrying more than one charge is usually not extensive beyond the first stage. Additional examples of the first stage in the ionization of hydrated metal ions are:

$\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{2+}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{a}}=2.74$
$\text{Cu}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{2+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Cu}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{\text{+}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{a}}=~6.3$
$\text{Zn}{\left({\text{H}}_{2}\text{O}\right)}_{4}{}^{2+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Zn}{\left({\text{H}}_{2}\text{O}\right)}_{3}\left({\text{OH}\right)}^{\text{+}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{a}}=9.6$

## Hydrolysis of [al(h 2 O) 6 ] 3+

Calculate the pH of a 0.10- M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion ${\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}\right]}^{3+}$ in solution.

## Solution

In spite of the unusual appearance of the acid, this is a typical acid ionization problem.

1. Determine the direction of change . The equation for the reaction and K a are:
$\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{\text{2+}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{a}}=1.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

The reaction shifts to the right to reach equilibrium.
2. Determine x and equilibrium concentrations. Use the table:
3. Solve for x and the equilibrium concentrations . Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:
${K}_{\text{a}}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{2+}\right]}{\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}\right]}$

$=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.10-x}\phantom{\rule{0.2em}{0ex}}=1.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

Solving this equation gives:
$x=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}M$

From this we find:
$\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=0+x=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}M$

$\text{pH}=\text{−log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\phantom{\rule{0.2em}{0ex}}=2.92\phantom{\rule{0.2em}{0ex}}\left(\text{an acidic solution}\right)$
4. Check the work . The arithmetic checks; when 1.2 $×$ 10 −3 M is substituted for x , the result = K a .

What is $\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{2+}\right]$ in a 0.15- M solution of Al(NO 3 ) 3 that contains enough of the strong acid HNO 3 to bring [H 3 O + ] to 0.10 M ?

2.1 $×$ 10 −5 M

The constants for the different stages of ionization are not known for many metal ions, so we cannot calculate the extent of their ionization. However, practically all hydrated metal ions other than those of the alkali metals ionize to give acidic solutions. Ionization increases as the charge of the metal ion increases or as the size of the metal ion decreases.

## Key concepts and summary

The characteristic properties of aqueous solutions of Brønsted-Lowry acids are due to the presence of hydronium ions; those of aqueous solutions of Brønsted-Lowry bases are due to the presence of hydroxide ions. The neutralization that occurs when aqueous solutions of acids and bases are combined results from the reaction of the hydronium and hydroxide ions to form water. Some salts formed in neutralization reactions may make the product solutions slightly acidic or slightly basic.

Solutions that contain salts or hydrated metal ions have a pH that is determined by the extent of the hydrolysis of the ions in the solution. The pH of the solutions may be calculated using familiar equilibrium techniques, or it may be qualitatively determined to be acidic, basic, or neutral depending on the relative K a and K b of the ions involved.

## Chemistry end of chapter exercises

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) Al(NO 3 ) 3

(b) RbI

(c) KHCO 2

(d) CH 3 NH 3 Br

Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) FeCl 3

(b) K 2 CO 3

(c) NH 4 Br

(d) KClO 4

(a) acidic; (b) basic; (c) acidic; (d) neutral

Novocaine, C 13 H 21 O 2 N 2 Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 $×$ 10 −6 . Is a solution of novocaine acidic or basic? What are [H 3 O + ], [OH ], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g/mL.

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