Question

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a solution that is (a) \(0.0062 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) and \(0.0105 \mathrm{M} \mathrm{BaCl}_{2} ;\) (b) \(0.315 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) and \(0.486 \mathrm{M} \mathrm{NH}_{3} ;\) (c) \(0.196 \mathrm{M} \mathrm{NaOH}\) and \(0.264 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}\)

Short answer

\([OH^-]\) in solution (a) is 0.0124M, in solution (b) is negligible, and in solution (c) is 0.196M.

Step-by-step solution

Identify the hydroxide ion contribution for each solution

(a) In the first case, barium hydroxide (\(Ba(OH)_2\)) dissociates into barium ions and two hydroxide ions per one molecule of \(Ba(OH)_2\). Therefore, the concentration of hydroxide ions is twice the concentration of barium hydroxide, which is \(2*(0.0062M)=0.0124M\). (b) The second case involves ammonia, which reacts with water to create hydroxide ions; however, the equilibrium constant for this process is small, so we assume that the concentration of hydroxide ions is negligible relative to those introduced by bases. Therefore, the \([OH^-]\) is negligible. (c) Sodium hydroxide (\(NaOH\)) is a strong base, which completely dissociates to give one hydroxide ion per one \(NaOH\) molecule. Therefore, we get a hydroxide ion concentration equal to the sodium hydroxide concentration, which is \(0.196M\).

Calculate \([OH^-]\) when a salt is introduced to the solution

When a salt is introduced to the solution, it may affect the \([OH^-]\). However, in cases (a) and (c), barium chloride and ammonium chloride do not contribute to \([OH^-]\) because they do not dissociate to produce hydroxide ions. Hence, the \([OH^-]\) for cases (a) and (c) remains 0.0124M and 0.196M respectively.

Final Calculation

To summarize, \([OH^-]\) in solution (a) is 0.0124M, in solution (b) is negligible, and in solution (c) is 0.196M.

Key concepts

Barium Hydroxide Dissociation

Barium hydroxide, known as \(Ba(OH)_2\), plays an essential role in solution chemistry. When it dissociates in water, each formula unit of \(Ba(OH)_2\) dissociates into one \(Ba^{2+}\) ion and two \(OH^-\) ions.
This means that the dissociation greatly impacts the [OH^-] concentration in a solution.
If you start with a \(0.0062M\) solution of \(Ba(OH)_2\), the concentration of hydroxide ions will be twice that of \(Ba(OH)_2\) due to the release of two hydroxide ions per molecule.

• Concentration of \(OH^-\) = \(2 \times 0.0062M = 0.0124M\).
• This understanding is crucial for calculating pH and other equilibrium aspects in chemistry.

It's crucial to remember that substances like \(BaCl_2\) do not affect hydroxide ion concentration in such scenarios, as they do not release \(OH^-\) ions upon dissociation.

Ammonia Equilibrium Constant

Ammonia (\(NH_3\)) is a weak base, known for its ability to partially ionize in water. This process is governed by its equilibrium constant \(K_b\), which measures the extent of ammonia's ionization to form \(OH^-\) ions and \(NH_4^+\) ions:
\(NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\).
The equilibrium constant, however, is small, indicating limited ionization.
In a solution with ammonia and, say, ammonium sulfate, the presence of ammonium ions (from a salt) can shift the balance back due to the common ion effect, further reducing the production of \(OH^-\) ions.

• Essentially, this often results in the [OH^-] being negligible in solutions with both ammonia and ammonium salts.
• Understanding this behavior is key to recognizing why ammonia's influence on [OH^-] is often minimal.

Hence, solutions containing both \((NH_4)_2SO_4\) and \(NH_3\)
often result in a negligible [OH^-] concentration, due to their competing equilibria.

Sodium Hydroxide Strong Base

Sodium hydroxide (\(NaOH\)) is renowned as a classic example of a strong base. This substance dissociates completely in water to produce \(Na^+\) and \(OH^-\) ions:
\(NaOH \rightarrow Na^+ + OH^-\).
The complete dissociation means that the hydroxide ion concentration directly correlates with the concentration of \(NaOH\).

• For example, a solution of \(0.196M\) \(NaOH\) will directly lead to an \(OH^-\) concentration of \(0.196M\).
• This straightforward relationship makes \(NaOH\) predictable and reliable when considering pH and alkalinity adjustments in solutions.

Sodium hydroxide's significant contribution to [OH^-] is particularly important in case scenarios where other chemicals don’t alter its dissociation.
This powerful impact underscores the nature of strong bases in chemical reactions and equilibria.