Problem 108 - Chemistry The Central Science

Question

(a) Write the net ionic equation for the reaction that occurs when a solution of hydrochloric acid (HCl) is mixed with a solution of sodium formate \(\left(\mathrm{NaCHO}_{2}\right) .\) (b) Calculate the equilibrium constant for this reaction. (c) Calculate the equilibrium concentrations of \(\mathrm{Na}^{+}, \mathrm{Cl}^{-}, \mathrm{H}^{+}, \mathrm{CHO}_{2}^{-},\) and \(\mathrm{HCHO}_{2}\) when \(50.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{HCl}\) is mixed with \(50.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{NaCHO}_{2}\)

Step-by-Step Solution

Answer

The net ionic equation for the reaction between HCl and NaCHO2 is: \[H^{+}(aq) + CHO_2^{-}(aq) \rightarrow HCHO_2(aq)\] The equilibrium constant, \(K_a\), for this reaction is 1.78×10^{-4}. The equilibrium concentrations are as follows: \[[Na^{+}] = 0.075M,\, [Cl^{-}] = 0.075M,\, [H^{+}] ≈ 0.0706M,\, [CHO_{2}^{-}] ≈ 0.0706M,\, [HCHO_2] = 4.45×10^{-3}M\]

Step 1: Write the balanced molecular equation

When hydrochloric acid (HCl) reacts with sodium formate (NaCHO2), the H+ ion from the acid reacts with the CHO2- ion from the sodium formate to form formic acid (HCHO2), while sodium (Na+) and chloride (Cl-) ions remain unchanged. This can be represented as: \[HCl(aq) + NaCHO_2(aq) \rightarrow HCHO_2(aq) + NaCl(aq)\]

Step 2: Write the balanced ionic equation

Separate the strong electrolytes into their respective ions: \[H^{+}(aq) + Cl^{-}(aq) + Na^{+}(aq) + CHO_2^{-}(aq) \rightarrow HCHO_2(aq) + Na^{+}(aq) + Cl^{-}(aq)\]

Step 3: Find the net ionic equation

Remove the spectator ions, i.e., the ions that don't change during the reaction: \[H^{+}(aq) + CHO_2^{-}(aq) \rightarrow HCHO_2(aq)\] This is the net ionic equation for the reaction.

Step 4: Calculate the equilibrium constant

As this is an acid-base reaction, we can write the equilibrium constant expression as \(K_a\), where: \[K_a = \frac{[HCHO_2]}{[H^{+}][CHO_2^{-}]}\] Given that the pKa value for formic acid is 3.75, we can calculate \(K_a\) using the relation: \[K_a = 10^{-pKa} = 10^{-3.75}\] Therefore, \(K_a = 1.78×10^{-4}\).

Step 5: Calculate the initial concentration of each ion

We are given that both solutions have a concentration of 0.15 M and a volume of 50 mL each. As the total volume is 100 mL, the initial concentrations will be: \[[H^{+}] = [Cl^{-}] = [Na^{+}] = [CHO_2^{-}] = 0.075M\] \[[HCHO_2] = 0\]

Step 6: Calculate the final concentrations using the equilibrium constant

Let x represent the change in concentration for all species during the reaction: \[[HCHO_2] = x\] \[[H^{+}] = 0.075 - x\] \[[CHO_2^{-}] = 0.075 - x\] Now, substitute the equilibrium expression: \[K_a = \frac{x}{(0.075-x)^2}\] Solve for x = 4.45×10^{-3}M Therefore, the equilibrium concentrations are: \[[HCHO_2] = 4.45×10^{-3}, [Na^{+}] = [Cl^{-}] = 0.075\] \[[H^{+}] = [CHO_2^{-}] = 0.075 - 4.45×10^{-3}\] So, the equilibrium concentrations are: \[[Na^{+}] = 0.075M,\, [Cl^{-}] = 0.075M,\, [H^{+}] ≈ 0.0706M,\, [CHO_{2}^{-}] ≈ 0.0706M,\, [HCHO_2] = 4.45×10^{-3}M\]

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the backward reaction, resulting in no overall change in the concentrations of reactants and products over time. This dynamic equilibrium is crucial for understanding the net ionic equations in acid-base reactions, where the dissociation and recombination of ions reach a balance.

To visualize this, imagine a chemical reaction as a busy intersection where cars are constantly moving back and forth between two streets. At chemical equilibrium, the number of cars entering each street is equal to the number of cars leaving. Similarly, in the reaction between hydrochloric acid and sodium formate, the formation rate of formic acid (the forward reaction) balances with the dissociation rate back to ions (the reverse reaction). It's essential to identify and focus on the ions that undergo change—these are represented in the net ionic equation.

Acid-Base Reactions

In acid-base reactions, an acid donates a proton (H⁺) to a base. The exercise example involves a common acid-base reaction where HCl, a strong acid, donates a proton to the formate ion (CHO₂^-), which acts as a base. This reaction forms formic acid, HCHO₂. In solutions, such reactions are often depicted by net ionic equations, which ignore spectator ions and show only the species directly involved in the reaction.

Understanding net ionic equations involves recognizing the role of water as an implicit participant—especially when dealing with acids and bases. Water provides the solvent environment necessary for the dissociation of ions and the formation of hydronium ions (H₃O⁺).

Equilibrium Constant Calculation

The equilibrium constant, represented as K_a for acid dissociation reactions, quantitatively describes the ratio of the concentration of the products to the reactants at equilibrium. It's a measure of how far the reaction proceeds before reaching equilibrium.

In the exercise, the K_a value is calculated from the known pKa value of formic acid using the formula K_a = 10^-pKa. This value is crucial for predicting the behavior of the acid in different environmental conditions and is also key to determining the equilibrium concentrations of the various species in solution. By performing a stoichiometric calculation, we arrive at the equilibrium concentrations, which adhere to the calculated K_a. It's important to apply rigorous mathematical treatment to obtain precise equilibrium concentrations, as this deepens the understanding of the reaction dynamics.