Question

Which of the following would form a buffer if added to $650.0\mathrm{mL}$ of $0.40M\mathrm{Sr}(\mathrm{OH}{)}_2?$ (a) $1.00\mathrm{mol}$ of $\mathrm{HF}$ (b) $0.75\mathrm{mol}$ of $\mathrm{HF}$ (c) $0.30\mathrm{mol}$ of $\mathrm{HF}$ (d) $0.30\mathrm{mol}$ of $\mathrm{NaP}$ (e) $0.30\mathrm{mol}$ of $\mathrm{HCl}$ Explain your reasoning in each case.

Short answer

(a) 1.00 mol of HF, (b) 0.75 mol of HF, (c) 0.30 mol of HF, (d) 0.30 mol of NaP, (e) 0.30 mol of HCl. Answer: (c) 0.30 mol of HF

Step-by-step solution

Identify the initial amount of the base in moles

First, calculate the number of moles of $\mathrm{Sr}(\mathrm{OH}{)}_2$ present in the solution. We can do this by multiplying the concentration by the volume. Number of moles of $\mathrm{Sr}(\mathrm{OH}{)}_2=$ Concentration $\times$ Volume $=0.40\mathrm{M}\times 0.650\mathrm{L}=0.26\mathrm{mol}$

Analyze each option

# (a) Adding $1.00\mathrm{mol}$ of $\mathrm{HF}$: Hydrofluoric acid is a weak acid, and its conjugate base is fluoride ion (${\mathrm{F}}^{-}$). However, the amount of $\mathrm{HF}$ is too large to form a buffer, as it would neutralize all the $\mathrm{Sr}(\mathrm{OH}{)}_2$ and leave an excess of $\mathrm{HF}$. This means that the solution would become acidic instead. (b) Adding $0.75\mathrm{mol}$ of $\mathrm{HF}$: Hydrofluoric acid is a weak acid and forms a conjugate base, (${\mathrm{F}}^{-}$). However, the amount of $\mathrm{HF}$ is still too large to form a buffer, as it would neutralize all the $\mathrm{Sr}(\mathrm{OH}{)}_2$ and leave an excess of $\mathrm{HF}$. This means that the solution would still become acidic. (c) Adding $0.30\mathrm{mol}$ of $\mathrm{HF}$: Hydrofluoric acid is a weak acid and forms a conjugate base, (${\mathrm{F}}^{-}$). As $0.30\mathrm{mol}$ would lead to some unreacted $\mathrm{Sr}(\mathrm{OH}{)}_2$ remaining in the solution after reaction with $\mathrm{HF}$, both weak acid and conjugate base are present in the solution. Therefore, this option can form a buffer. (d) Adding $0.30\mathrm{mol}$ of $\mathrm{NaP}$: $\mathrm{NaP}$ is not a suitable candidate to form a buffer because it is the salt of a strong base and strong acid, which doesn't create a weak conjugate acid/base. (e) Adding $0.30\mathrm{mol}$ of $\mathrm{HCl}$: Hydrochloric acid is a strong acid and forms its conjugate base, chloride ion (${\mathrm{Cl}}^{-}$), which will not form a buffer with a strong base like $\mathrm{Sr}(\mathrm{OH}{)}_2$.

Identify the correct option

Based on the analysis in step 2, we can conclude that option (c) which is adding $0.30\mathrm{mol}$ of $\mathrm{HF}$, would form a buffer when added to $650.0\mathrm{mL}$ of $0.40M\mathrm{Sr}(\mathrm{OH}{)}_2$.

Key concepts

Buffer Chemistry

Understanding buffer chemistry is essential for students tackling problems related to chemical equilibrium in solutions. A buffer solution resists changes in pH when small amounts of acid or base are added. This stability is due to the presence of both a weak acid and its conjugate base, or a weak base and its conjugate acid, in significant quantities.

For a solution to act as a buffer, it must contain species capable of reacting with any added hydroxide ions (OH^-) or hydronium ions (H₃O⁺) to maintain the pH. A classic example involves a weak acid, like acetic acid, and its conjugate base, the acetate ion, coming from its salt, such as sodium acetate.

When a strong acid is introduced to this mixture, the acetate ion reacts to form more acetic acid, minimizing pH fluctuation. Similarly, when a strong base is added, the acetic acid donates a proton to form water and more acetate ions, again protecting the pH from significant change.

Weak Acids and Bases

To delve deeper into buffer solutions, it's important to understand the behavior of weak acids and bases. Unlike their strong counterparts that dissociate completely in water, weak acids and bases only partly dissociate, establishing an equilibrium between the undissociated species and the resulting ions.

Weak acids, such as hydrofluoric acid (HF), only donate a fraction of their protons to the solution, resulting in an equilibrium mixture of HF and its conjugate base (F^-). Similarly, weak bases only accept some protons from water, leading to a dynamic equilibrium. The presence of both the weak acid/base and its conjugate base/acid is what allows the solution to minimize changes in pH and is why these substances are key components in buffer solutions.

Stoichiometry

Stoichiometry plays a crucial role in creating buffer solutions. It involves the quantitative aspects of chemical reactions and compounds, determining the exact amounts of reactants needed for a reaction to proceed without any limiting reagents.

In the context of the textbook exercise, stoichiometry is applied to ensure that an appropriate mole ratio between the weak acid, HF, and the strong base, Sr(OH)₂, is established. Too much of the weak acid would completely neutralize the strong base, leading to an acidic solution rather than a buffer. However, when the stoichiometric amount results in unreacted weak acid and the formation of the weak acid's conjugate base, a buffer solution is successfully created. This demonstrates the importance of balancing chemical equations and measuring reactant amounts to form a desired product, in this case, a buffer solution.