KEY2CHEM

Solutions of Weak Acids and Bases

 

Aqueous solutions of salts can be acidic, basic, or neutral, depending on the ions in the solution. Neutral solutions are formed when both the cation and the anion are spectator ions (anions of strong acids or cations of strong bases). These spectator ions will not undergo proton transfer reactions with water, resulting in a neutral solution.

An acidic solution is one where the cation acts as a weak acid (increasing \([\require{mhchem}\ce{H3O+}]\)). This cation can be a small, highly charged metal cation, such as \(\require{mhchem}\ce{Al ^{3+}}\) or the cation of a weak base, such as \(\require{mhchem}\ce{NH4+}\). These ions undergo proton transfer with water to increase\( [\require{mhchem}\ce{H3O+}]\): \(\require{mhchem}\ce{NH4+(aq) + H2O(l) <=> H3O+(aq) + NH3(aq)}\).

A basic solution is generated when the salt contains the anion of a weak acid (increasing\( [\require{mhchem}\ce{OH-}]\)). Examples are \(\require{mhchem}\ce{F-}\) and \(\require{mhchem}\ce{CH3COO-}\).


Example 1.

 

Which ion is a spectator ion?

A. \(\require{mhchem}\ce{NH3}\)

B. \(\require{mhchem}\ce{Cl-}\)

C. \(\require{mhchem}\ce{F-}\)

 

 

 

Solution

 

B. \(\require{mhchem}\ce{Cl-}\)

\(\require{mhchem}\ce{HCl}\) is a strong acid (undergoes complete proton transfer in water), so its anion (\(\require{mhchem}\ce{Cl-}\)) is a negligible base (a spectator ion). \(\require{mhchem}\ce{NH3}\) and \(\require{mhchem}\ce{F-} \) both act as weak bases in aqueous solutions.

 


Example 2.

 

Which solution will have \(pH < 7\)?

 

A. \(\require{mhchem}\ce{NaF}\)

B. \(\require{mhchem}\ce{NaNO3}\)

C. \(\require{mhchem}\ce{NH4NO3}\)

 

 

Solution

 

C. \(\require{mhchem}\ce{NH4NO3}\)

Since \(\require{mhchem}\ce{NO3-}\) is the anion of a strong acid (\(\require{mhchem}\ce{HNO3}\)), it is a spectator ion. Since \(\require{mhchem}\ce{Na+}\) is the cation of a strong base (\(\require{mhchem}\ce{NaOH}\)), it is also a spectator ion. Since \(\require{mhchem}\ce{F-}\) is the anion of a weak acid (\(\require{mhchem}\ce{HF}\)), it acts as a weak base, yielding a solution with \(pH > 7\). Since \(\require{mhchem}\ce{NH4+}\) is the cation of a weak base (\(\require{mhchem}\ce{NH3}\)), it acts as a weak acid, yielding a solution with \(pH < 7\).


Example 3.

What is the pH of a \(0.1 \text{ M}\) solution of\( \require{mhchem}\ce{CH3COONa}\) at \(25^\circ C\)? The \(K_a\) of \(\require{mhchem}\ce{CH3COOH}\) is \(1.8 \times 10 ^{-5}\).

 

 

A. 7.00

B.  5.13

C. 8.87

 

 

 

 

 

 

Solution

 

C. 8.87

 

 

Since \(\require{mhchem}\ce{Na+}\) is a spectator ion, the pH is determined by \(\require{mhchem}\ce{CH3COO-}\), which acts as a weak base:

\(\require{mhchem}\ce{CH3COO^{-}(aq) + H2O(l) <=> CH3COOH(aq) + OH^{-}(aq)}\)

The equilibrium constant describing this equation is \(K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10 ^{-14}}{1.8 \times 10 ^{-5}} = 5.6 \times 10 ^{-10}\). Let \(x =\) change in concentration. Because \(x\) is small relative to the initial \([\require{mhchem}\ce{CH3COO-}]\), substituting the change values yields:

\(K_b = \frac{x^2}{0.1} = 5.6 \times 10 ^{-10}\)

and \(x = 7.48 \times 10 ^{-6}\)

Since \(x = [\require{mhchem}\ce{OH-}]\), the \(pOH = -log[\require{mhchem}\ce{OH-}] = -log(7.48 \times 10 ^{-6}) = 5.13\)

\(pH + pOH = 14.00, \text{ so } pH = 14.00 – pOH = 14.00 - 5.13 = 8.87\)