G12 C3 Chemical Equilibrium Short Question Answers

Chapter 3: Chemical Equilibrium

Short Question Answer

i. Explain the common ion effect with a suitable example.

The **common ion effect** is the suppression of the ionization of a weak electrolyte by the addition of a strong electrolyte containing a common ion. According to Le Châtelier's principle, the addition of a common ion shifts the equilibrium of the weak electrolyte to the left, decreasing its ionization.

Example: Consider a solution of acetic acid ($$CH_3COOH$$), a weak acid. Its ionization equilibrium is:

$$CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)$$

If we add a strong electrolyte like sodium acetate ($$CH_3COONa$$), which fully dissociates into $$CH_3COO^-$$ and $$Na^+$$, the concentration of the common ion, $$CH_3COO^-$$, increases. This increase shifts the equilibrium to the left, causing less $$CH_3COOH$$ to ionize and suppressing the production of $$H^+$$, thus increasing the pH of the solution.

ii. Differentiate between strong and weak acids.

• **Strong Acids:** These acids **ionize completely** in an aqueous solution. Their reactions are considered irreversible and are represented by a single arrow ($$\rightarrow$$). Examples include $$HCl$$ and $$H_2SO_4$$. Their dissociation constant ($$K_a$$) is very large.
• **Weak Acids:** These acids **ionize partially** in an aqueous solution, meaning the reaction reaches an equilibrium. They are represented by a double arrow ($$\rightleftharpoons$$). Examples include $$CH_3COOH$$ and $$H_2CO_3$$. Their dissociation constant ($$K_a$$) is very small.

iii. Differentiate between strong and weak bases.

• **Strong Bases:** These bases **dissociate completely** in an aqueous solution. The reaction is irreversible ($$\rightarrow$$). Examples include $$NaOH$$ and $$KOH$$. Their dissociation constant ($$K_b$$) is very large.
• **Weak Bases:** These bases **ionize partially** in an aqueous solution, establishing an equilibrium ($$\rightleftharpoons$$). Examples include $$NH_3$$ and $$CH_3NH_2$$. Their dissociation constant ($$K_b$$) is very small.

iv. Define a buffer solution and provide an example.

A **buffer solution** is a solution that resists changes in pH upon the addition of a small amount of an acid or a base. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid.

Example: A buffer can be made by mixing a weak acid like **acetic acid ($$CH_3COOH$$)** with its conjugate base, **sodium acetate ($$CH_3COONa$$)**.

v. How does a buffer solution control pH?

A buffer solution contains a weak acid and its conjugate base. When a small amount of acid is added, the conjugate base neutralizes it. When a small amount of base is added, the weak acid neutralizes it. This prevents a significant change in pH.

Chemical Equations for an acetic acid/acetate buffer:

When an acid ($$H^+$$) is added:

$$CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)$$

The acetate ion ($$CH_3COO^-$$) reacts with the added $$H^+$$ to form a weak acid, keeping the pH stable.

When a base ($$OH^-$$) is added:

$$CH_3COOH(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + H_2O(l)$$

The acetic acid ($$CH_3COOH$$) reacts with the added $$OH^-$$ to form water and the conjugate base, keeping the pH stable.

vi. Describe the uses of buffer solutions.

Buffer solutions have wide-ranging applications:

• **Biological Systems:** The most critical example is blood, which is a natural buffer system. It maintains the pH between 7.35 and 7.45, essential for enzyme function and overall health.
• **Pharmaceuticals:** Buffers are used to maintain the pH of drug formulations, ensuring their stability and effectiveness.
• **Food and Beverage Industry:** They are used to control the pH of products like soft drinks, jams, and jellies to prevent microbial growth and maintain flavor.
• **Chemical Analysis:** Buffers are crucial in laboratory settings to control the pH of a reaction for accurate results, particularly in chromatography and titration.

vii. Explain how HCO₃⁻ plays a role in controlling pH in the blood.

The bicarbonate buffer system is the primary mechanism for regulating blood pH. It consists of carbonic acid ($$H_2CO_3$$) and its conjugate base, the bicarbonate ion ($$HCO_3^-$$).

• If blood pH starts to drop (becomes acidic), the bicarbonate ion ($$HCO_3^-$$, the weak base) reacts with excess $$H^+$$ to form carbonic acid, which is then converted into $$CO_2$$ and exhaled by the lungs.

  $$HCO_3^-(aq) + H^+(aq) \rightleftharpoons H_2CO_3(aq) \rightleftharpoons H_2O(l) + CO_2(g)$$

• If blood pH starts to rise (becomes basic), carbonic acid ($$H_2CO_3$$), formed from $$CO_2$$ and $$H_2O$$, donates a proton to react with the excess $$OH^-$$ ions, maintaining the pH.

  $$H_2CO_3(aq) + OH^-(aq) \rightleftharpoons HCO_3^-(aq) + H_2O(l)$$

The ability of the bicarbonate ion to neutralize excess acid and carbonic acid to neutralize excess base is what keeps the blood pH within its narrow, life-sustaining range.

viii. Calculate the concentration of a slightly soluble salt given its solubility product constant ($$K_{sp}$$).

To calculate the concentration, you first write the dissolution equation for the salt. For a generic salt, $$A_xB_y$$, the equilibrium is:

$$A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)$$

The solubility product constant ($$K_{sp}$$) is defined as:

$$K_{sp} = [A^{y+}]^x [B^{x-}]^y$$

Let 's' be the molar solubility (concentration) of the salt. At equilibrium, the concentrations of the ions will be:

• $$[A^{y+}] = xs$$
• $$[B^{x-}] = ys$$

Substitute these into the $$K_{sp}$$ expression:

$$K_{sp} = (xs)^x (ys)^y = x^x y^y s^{x+y}$$

To find the concentration 's', you rearrange the equation to solve for s:

$$s = \sqrt[x+y]{\frac{K_{sp}}{x^x y^y}}$$

By plugging in the given $$K_{sp}$$ value and the stoichiometric coefficients (x and y) of the salt, you can calculate its molar concentration (solubility).

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