Problem 110 - 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 (NaCHO \(_{2} )\) . (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 \(\mathrm{MCl}\) is mixed with 50.0 \(\mathrm{mL}\) of 0.15 \(\mathrm{MNaCHO}_{2} .\)

Step-by-Step Solution

Answer

The net ionic equation for the reaction of HCl with NaCHO2 is: H+(aq) + CHO2^-(aq) -> HCHO2(aq) The equilibrium constant, K, is calculated to be 5.56 x 10^3. The equilibrium concentrations of Na+, Cl-, H+, CHO2^-, and HCHO2 are approximately 0.0001 M, 0.150 M, 0.0001 M, 0.0001 M, and 0.0749 M, respectively.

Step 1: Write the balanced chemical equation

The balanced chemical equation for the reaction of HCl (a strong acid) with NaCHO2 (a salt containing a weak acid) can be written as: HCl(aq) + NaCHO2(aq) -> NaCl(aq) + HCHO2(aq)

Step 2: Write the net ionic equation

Strong electrolytes like strong acids (HCl), strong bases, and soluble salts dissociate completely into ions in solution. So, we can write the net ionic equation for the reaction as: H+(aq) + CHO2^-(aq) -> HCHO2(aq)

Step 3: Calculate the equilibrium constant

The equilibrium constant, K, for any reaction can be written in terms of the concentrations of the products divided by the concentrations of the reactants, with each concentration raised to the power of the stoichiometric coefficients in the balanced net ionic equation: K = \(\frac{[HCHO2]}{[H^+][CHO2^-]}\) This reaction produces the conjugate acid (HCHO2) of the weak base (CHO2^-). Since the equilibrium constant expression is in terms of the weak acid dissociation constant (Ka), we can find K by using the relationship: K = \(\frac{1}{Ka}\) To find the value of Ka, use a reference for the Ka of the conjugate parent acid, formic acid (HCOOH or HCHO2): Ka = 1.8 x 10^-4 (for HCHO2) Now, we can find K: K = \(\frac{1}{1.8 \times 10^{-4}}\) = 5.56 x 10^3

Step 4: Calculate the initial concentrations

The initial concentrations can be determined by using the volumes and molarities provided: Initial moles of HCl = (50.0 mL)(0.15 M) = 7.5 mmol Initial moles of NaCHO2 = (50.0 mL)(0.15 M) = 7.5 mmol The total volume of the solution is 100 mL. Initial concentration of H+ = \(\frac{7.5\,\text{mmol}}{100\,\text{mL}}\)= 0.075 M Initial concentration of CHO2- = \(\frac{7.5\,\text{mmol}}{100\,\text{mL}}\)= 0.075 M The reaction consumes equal moles of H+ and CHO2-, and produces equal moles of HCHO2: [H+] = 0.075 M - x [CHO2^-] = 0.075 M - x [HCHO2] = x

Step 5: Find the equilibrium concentrations

To find the equilibrium concentrations, we can plug the expressions for the concentrations into the equilibrium constant expression and solve for x: K = \(\frac{x}{(0.075-x)(0.075-x)}\) 5.56 x 10^3 = \(\frac{x}{(0.075-x)^2}\) Solving for x, we get about 0.0749 M. Now that we have x, we can find the equilibrium concentrations of each species: [Na+]=[H+] = 0.075 - 0.0749 ≈ 0.0001 M [Cl-] = 0.150 M (since the initial concentration of HCl was 0.15 M and the final concentration of H+ is about 0.0001 M) [H+] = 0.0001 M [CHO2^-] = 0.0001 M [HCHO2] = 0.0749 M Therefore, the equilibrium concentrations of Na+, Cl-, H+, CHO2^-, and HCHO2 are approximately 0.0001 M, 0.150 M, 0.0001 M, 0.0001 M, and 0.0749 M, respectively.

Key Concepts

Equilibrium Constant

The **equilibrium constant**, denoted as \( K \), is a value that helps quantify the balance of concentrations between products and reactants in a reversible chemical reaction at equilibrium. It gives insight into the favorability of a reaction in either direction.

Imagine a see-saw that balances perfectly for a chemical reaction. If it tilts towards the products, it indicates that the equilibrium constant is large. If it tilts towards reactants, the constant is smaller.

For the net ionic reaction in our problem, \[ H^+(aq) + CHO_2^-(aq) \rightleftharpoons HCHO_2(aq) \], the equilibrium constant \( K \) is calculated using the formula:

• \( K = \frac{[HCHO_2]}{[H^+][CHO_2^-]} \)

Here, \([HCHO_2]\) is the concentration of the product, and \([H^+]\) and \([CHO_2^-]\) are the concentrations of the reactants. A high \( K \) value signifies that at equilibrium, the products are more prevalent compared to reactants.

Chemical Reaction

A **chemical reaction** involves the rearrangement of atoms to transform reactants into products, often involving energy changes. In this exercise, when hydrochloric acid \((HCl)\) meets sodium formate \((NaCHO_2)\), they interact to form sodium chloride \((NaCl)\) and formic acid \((HCHO_2)\).

The balanced chemical equation for this is:

• \( HCl(aq) + NaCHO_2(aq) \rightarrow NaCl(aq) + HCHO_2(aq) \)

In any balanced equation, the number of atoms for each element is equal on both sides. The shift in this equation occurs because the chloride ion \((Cl^-)\) from \( HCl \) combines with the sodium ion \((Na^+)\) from \( NaCHO_2 \), while the hydrogen ion \((H^+)\) combines with the formate ion \((CHO_2^-)\) to create formic acid.

Every chemical reaction has its own special "fingerprint" portrayed in its balanced equation, allowing chemists to predict and quantify how substances will transform.

Weak Acid

A **weak acid** is an acid that does not fully dissociate in solution, unlike strong acids that completely separate into ions. Formic acid, represented in our reaction as \( HCHO_2 \), is a classic example of a weak acid.

When formic acid dissolves in water, only a small fraction of its molecules donate protons \((H^+)\) to form the ions \( H^+ \) and \( CHO_2^- \). Weak acids have characteristic equilibrium constants, called **dissociation constants** \( (K_a) \), that help us understand this partial dissociation:

• \( K_a = \frac{[H^+][CHO_2^-]}{[HCHO_2]} \)

In the context of our exercise, the \( K_a \) value for formic acid is \( 1.8 \times 10^{-4} \), indicating that the acid remains largely undissociated. This underscores the fact that weak acids tend to remain mostly as molecules rather than ions in solution.

Understanding weak acids helps reveal the nuanced dance of molecules as they partially transform in water.

Equilibrium Concentrations

**Equilibrium concentrations** represent the amounts of each species (reactants and products) present in a reaction mixture once the system has reached equilibrium. At this point, the rates of the forward and reverse reactions are equal, creating a stable mix that doesn’t change over time without external influence.

In our exercise, we calculated these concentrations based on initial conditions of 50.0 mL solutions each of 0.15 M \( HCl \) and \( NaCHO_2 \):

• \([Na^+] = [H^+] = 0.0001 \text{ M} \)
• \([Cl^-] = 0.150 \text{ M} \)
• \([CHO_2^-] = 0.0001 \text{ M} \)
• \([HCHO_2] = 0.0749 \text{ M} \)

These concentrations result from the establishment of equilibrium, where initial reactant concentrations were reduced to form products until balance was achieved.

Understanding equilibrium concentrations is key to predicting how a reaction will behave under various circumstances, and manipulating conditions to drive more products or reactants as needed.