Question

Using appropriate equilibrium constants but without doing detailed calculations, determine whether a solution can be simultaneously: (a) \(0.10 \mathrm{M} \mathrm{NH}_{3}\) and \(0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl},\) with \(\mathrm{pH}=6.07\) (b) \(0.10 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(0.058 \mathrm{M} \mathrm{HI}\) (c) \(0.10 \mathrm{M} \mathrm{KNO}_{2}\) and \(0.25 \mathrm{M} \mathrm{KNO}_{3}\) (d) \(0.050 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) and \(0.65 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (e) \(0.018 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) and \(0.018 \mathrm{M} \mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{COO}\) with \(\mathrm{pH}=4.20\) (f) \(0.68 \mathrm{M} \mathrm{KCl}, 0.42 \mathrm{M} \mathrm{KNO}_{3}, 1.2 \mathrm{M} \mathrm{NaCl},\) and \(0.55 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) with \(\mathrm{pH}=6.4\)

Short answer

Part (a), (b), and (d) cannot coexist with the given constraints, while part (c), (e), and (f) can.

Step-by-step solution

Analyze Component (a)

For the solution to be \(0.10 \, \mathrm{M} \, \mathrm{NH}_3\), and \(0.10 \, \mathrm{M} \, \mathrm{NH}_4 \mathrm{Cl}\) simultaneously with pH = 6.07, it should be noted that ammonia acts as a weak base (\(NH_3 \rightarrow NH_4^+ + OH^-\)), and \(NH_4Cl\) (ammonium chloride) gives \(NH_4^+\) which acts as a weak acid (\(NH_4^+ \rightarrow NH_3 + H^+\)). As the ratio of [\(NH_3\)] / [\(NH_4^+\)] equals 1 and the pKb of ammonia equals 4.75, a pH of 6.07 calculated from this buffer pair will not be possible as the pH is determined by \(pH = pKw - pKb + log ([NH_4^+]/[NH_3]) = 14 - 4.75 + log (1) = 9.25\). Therefore, the components could not coexist.

Analyze Component (b)

In this case, \(0.10 \, \mathrm{M} \, \mathrm{NaC}_2 \mathrm{H}_3 \mathrm{O}_2\) is a salt of a weak acid and will form a weak base (\(C_2H_3O_2^-\)). However, \(0.058 \, \mathrm{M} \, \mathrm{HI}\) is a strong acid that will significantly lower the pH and prevent the weak base from thriving. Hence, the components cannot coexist in the same solution.

Analyze Component (c)

For the case of \(0.10 \, \mathrm{M} \, \mathrm{KNO}_2\) and \(0.25 \, \mathrm{M} \, \mathrm{KNO}_3\), the salts of weak acids will create a weak base when dissolved. However, the existence of \(0.25 \, \mathrm{M} \, \mathrm{KNO}_3\), which is a strong electrolyte, does not affect the pH. Hence, these components could hypothetically coexist in the same solution without conflicting effects on pH.

Analyze Component (d)

\(0.050 \, \mathrm{M} \, \mathrm{Ba(OH)}_2\) is a strong base and \(0.65 \, \mathrm{M} \, \mathrm{NH}_4 \mathrm{Cl}\) is a salt of a weak acid, which will form a weak base. Since we have a strong base in the solution, it will significantly raise the pH and dominate over the weak base. Therefore, these components cannot coexist in the same solution.

Analyze Component (e)

In this case, both \(0.018 \, \mathrm{M} \, \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\) and \(0.018 \, \mathrm{M} \, \mathrm{NaC}_6 \mathrm{H}_5 \mathrm{COO}\) form a buffer system. Regardless of the pH level, these components can coexist in the same solution.

Analyze Component (f)

All the salts listed are either strong electrolytes (KCl and KNO3) or weak acids/bases (NaCl and \(NaCH_3 \mathrm{COO}\)) which individually do not change the pH. Therefore, if the pH value is manipulated to be 6.4 by adding an acid/base, these components can coexist.

Key concepts

Understanding Equilibrium Constants

Equilibrium constants are fundamental to the study of chemical reactions and chemistry itself. They give us a measure of the extent to which a chemical reaction occurs and, ultimately, when a system reaches a state of balance between reactants and products.

When a chemical reaction is reversible, reaching a point where the rate of the forward reaction equals the rate of the reverse reaction, it's said to be at equilibrium. The equilibrium constant, represented as K_eq, is the ratio of product concentrations to reactant concentrations at this point, each raised to the power of their respective coefficients from the balanced equation.

Calculating Equilibrium Constants

Consider a generic reaction where aA + bB ⇌ cC + dD; the equilibrium constant (K_eq) can be expressed as follows: \[ K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
where [A], [B], [C], and [D] are the molar concentrations of the reactants and products at equilibrium.

In the context of the exercise, the equilibrium constant helps students determine the feasibility of different ions coexisting in a solution. For instance, understanding how the different values of K_eq for ammonia and ammonium chloride impact the pH can demonstrate why certain combinations of chemicals in a solution may not be practically viable.

pH Calculation Made Simple

pH is a scale used to specify the acidity or basicity of an aqueous solution. In chemistry, pH calculation is a vital skill as it helps determine the hydrogen ion concentration ([H⁺]) in a solution, influencing the chemical behavior within that solution.

The pH scale typically ranges from 0 to 14, with 7 being neutral. A pH less than 7 indicates acidity, while a value greater than 7 indicates basicity. The pH is calculated using the following equation: \[ pH = -\log[H^+] \]
One key application of pH is in buffer solutions, which resist changes in pH upon the addition of an acid or base. In the textbook exercise, when students are asked to determine if certain solutions can coexist at a specific pH, they are being challenged to apply their understanding of pH in predicting the behavior of weak acids/bases and their conjugate pairs, which is crucial for creating buffer solutions.

An example from the exercise shows a scenario where ammonia and ammonium have an equal concentration, but due to the pK_b of ammonia, the resulting pH would not match the expected value. This discrepancy is something students need to recognize and understand through pH calculations.

The Role of Buffer Solutions

Buffer solutions are an extraordinary concept in chemistry, serving as a cloak of stability against pH changes in a solution. They are particularly made up of a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid. This pairing allows the solution to 'buffer' against substantial alterations in pH when small amounts of an acid or base are added.

How Buffers Work

Here's a simplified view: when you add an acid to a buffer, the conjugate base present in the buffer neutralizes the added H⁺ ions. Conversely, if a base is added, the weak acid in the buffer neutralizes the OH^- ions.

The capacity of a buffer to maintain pH is governed by the Henderson-Hasselbalch equation: \[ pH = pKa + \log \left( \frac{[A^-]}{[HA]} \right) \]
where pKa is the acid dissociation constant, and [A^-] and [HA] are the concentrations of the conjugate base and weak acid, respectively. In the textbook exercise provided, part (e) is a classical example of a buffer system where benzoic acid and its salt sodium benzoate are in equal concentrations, demonstrating the potential to maintain a stable pH of 4.20, which showcases the buffering action.