Problem 237 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(50 \mathrm{ml}\) of \(0.2 \mathrm{M} \mathrm{NaCN}\) is mixed with \(50 \mathrm{ml}\) of \(0.2\) M \(\mathrm{HCl}\), then \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{CN}^{-}\right]=\mathrm{x} \times 10^{-6}\) where \(\mathrm{x}\) is \(\left(\mathrm{K}_{b}\right.\) for \(\left.\overrightarrow{C N}^{-}=2 \times 10^{-5}\right)\)

Step-by-Step Solution

Answer

The value of \(x\) is approximately 7.07.

Step 1: Calculate Moles of Reactants

Calculate the moles of each reactant: NaCN: \( 50\, \text{ml} \times 0.2\, \text{M} = 0.01\, \text{moles} \) HCl: \( 50\, \text{ml} \times 0.2\, \text{M} = 0.01\, \text{moles} \)

Step 2: Determine Limiting Reactant

Both NaCN and HCl are present in equal moles (0.01 moles each); therefore, they fully neutralize each other.

Step 3: Reaction and Equilibrium Expression

NaCN reacts with HCl:\[ \text{NaCN} + \text{HCl} \rightarrow \text{NaCl} + \text{HCN} \]The reaction produces 0.01 moles of HCN in a total solution volume of \(100\, \text{ml} = 0.1\, \text{L} \). The concentration of HCN is \([\text{HCN}] = \frac{0.01}{0.1} = 0.1\, \text{M} \).

Step 4: Use Equilibrium to Find [H3O+],[CN-]

HCN partially ionizes: \[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]The equilibrium constant \(K_a\) is found using \(K_w = 10^{-14}\) and \(K_b = 2 \times 10^{-5}\): \[ K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10} \]Apply the equilibrium expression: \[ K_a = \frac{[H^+][CN^-]}{[HCN]} \]Given \([HCN] = 0.1\, \text{M}\), assume \([H^+] = [CN^-] = x \times 10^{-6}\):\[ 5 \times 10^{-10} = \frac{(x \times 10^{-6})^2}{0.1} \]Solve for \(x\).

Step 5: Solve the Equation

From the equation:\[ 5 \times 10^{-10} \times 0.1 = x^2 \times 10^{-12} \]\[ x^2 = 5 \times 10^{-11} \]\[ x = \sqrt{5 \times 10^{-11}} \]\[ x = 7.07 \]

Step 6: Final Calculation and Answer Verification

Therefore, \([H_3O^+]=[CN^-] = 7.07 \times 10^{-6}\, \text{M}\). Consequently, \(x = 7.07\).

Key Concepts

Acid-base reactions

Acid-base reactions are a subclass of chemical reactions involving the transfer of protons. An acid donates a proton ( H^+ ), whereas a base accepts a proton. When you mix solutions like NaCN and HCl , as in the given exercise, an acid-base reaction occurs.
The reaction can be summarized as follows:

• Acid: The HCl (hydrochloric acid) releases a proton to form Cl^- (chloride ion) and H^+ .
• Base: The NaCN accepts the proton, transforming into HCN (hydrocyanic acid) and Na^+ (sodium ion).

As both reactants are mixed in equal molar amounts in this specific scenario, they completely neutralize each other. This means the reaction goes to completion until the limited amount of the reactants has been entirely consumed. Therefore, the entire solution ends up containing new compounds, specifically, NaCl and HCN . These outcomes are significant because they further define the equilibrium properties. Understanding this transfer process is key to predicting the natural progression of the reaction when varying concentrations of acids and bases come into play.

Equilibrium constant

Chemical equilibrium deals with the state in which the concentrations of reactants and products remain steady over time. In the case of weak acids and bases that only partially ionize in water, an equilibrium is established between ionized and unionized species. The equilibrium constant (K_a for acids and K_b for bases) helps express this balance quantitatively.
In the given scenario, after NaCN and HCl completely react to form HCN, the hydrocyanic acid itself reaches an equilibrium:\[\text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^-\]The exercise demonstrates how to calculate K_a using the ion product of water (K_w = 10^{-14}) and the K_b known for CN^-:

• \[ \text{K}_a = \frac{K_w}{K_b} \]
• \[ \text{K}_a = \frac{10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10} \]

This constant helps determine the concentration of ions such as [H^+] and [CN^-] at equilibrium, reflecting the behavior of weak acids in aqueous solutions.

Limiting reactant calculation

The concept of the limiting reactant explains which reactant will be consumed first and limit the extent of a chemical reaction. Simply put, it's the "bottleneck" of a reaction.
In any mixture of chemicals, the limiting reactant is the one in the smaller stoichiometric amount compared to what the reaction requires.
In the exercise, both NaCN and HCl have the same initial molar amounts of 0.01 moles each. Given the stoichiometry of a 1:1 reaction ratio, they completely neutralize each other:\[\text{NaCN} + \text{HCl} \rightarrow \text{NaCl} + \text{HCN}\]Since neither exceeds the required amount to react, both are limiting reactants. This mutual limitation means the reaction proceeds until both reactants are exhausted, leaving no excess. Consequently, knowing which reactant limits the reaction extent allows for predicting the maximum amounts of products formed. This knowledge is crucial for analyzing the results from chemical reactions and influencing factors such as solution concentration and yield.