Chemical Equilibria

To find the solubility of in 0.10 M KI(aq), we start with the dissolution equation:

The solubility product ( ) is given as . When dissolves, let the solubility be . In a 0.10 M KI solution, the concentration of is increased by the dissociation of KI to 0.10 M. Therefore, the equilibrium concentrations are and since is small.

Substitute into the expression:

Solving for :

Answer:

To solve for , we use the Henderson-Hasselbalch equation:



Given:
pH = 4.80
pK of CH COOH = 4.75

Substitute the values:
4.80 = 4.75 + \log\left(\frac{[CH_3COO^-]}{[CH_3COOH]}\right)

Solve for :


Convert the logarithm to a ratio:

Calculate :


Thus, can be determined:


Substitute the calculated ratio:


Round to three decimal places: **0.122**

Isoelectric Point of Glutamic Acid

The isoelectric point ( ) of an amino acid is the pH at which the molecule carries no net electric charge. For glutamic acid, the equation to find involves averaging the values of the acidic groups:

Step 1: Identify the relevant values for deprotonation:

• (α-carboxyl group) = 2.19
• (side-chain carboxyl group) = 4.25

Step 2: Compute the average of these values for :

The equation for is:

Step 3: Verify the result:

The computed value of 3.22 is within the specified range of 3.21 to 3.23, confirming its correctness.

Final Answer:

Thus, the isoelectric point of glutamic acid, rounded to two decimal places, is 3.22.

To solve this problem, we need to determine how much additional water should be added to a weak acid solution to double its degree of dissociation.

Given:

• Initial concentration of HA,
• Degree of dissociation initially,
• Dissociation constant,

The degree of dissociation, , for a weak acid is given by:

Initial degree of dissociation is:

We want to double the degree of dissociation:

The final concentration of the acid after dilution should make the new degree of dissociation equal to :

Squaring both sides gives:

The concentration changes with dilution. Using the dilution formula:

The volume of water to be added is:

Since we need to choose the closest option: 300 mL is the correct volume to add.

Therefore, the correct answer is 300 mL.

We are given that the concentration of the acid is [HA] = 0.1M and the pH of the solution is 2.0. We need to find the pK_a of the acid.

From the pH, we can calculate the concentration of H⁺ ions using:

Next, we apply the acid dissociation expression for a weak acid HA:

HA \rightleftharpoons H^+ + A^- ,and use the formula for the acid dissociation constant K_a:

.

Assuming that the concentration of A⁻ formed is equal to the concentration of H⁺, we have:

Now, to find the pK_a:

Thus, the pK_a of the acid is approximately 2.9.

• (A) A catalyst only affects the rate of reaction, not the equilibrium position. Therefore, the introduction of a catalyst does not increase product formation at equilibrium.
• (B) For an exothermic reaction, increasing the temperature shifts the equilibrium toward the reactants (Le Chatelier’s Principle). Thus, the equilibrium constant decreases with an increase in temperature.
• (C) The equilibrium constant K_p is not directly affected by pressure for reactions involving gases unless there is a change in the number of moles of gas. Since this reaction has no change in the number of moles of gas, K_p remains unaffected by pressure.
• (D) According to Le Chatelier’s Principle, reducing the volume of a reaction vessel increases the pressure and shifts the equilibrium toward the side with fewer moles of gas. In this case, the product side (2 moles of H₂O) has fewer moles of gas than the reactant side (3 moles of gas), so decreasing the volume increases product formation.

Thus, the correct answers are (B) and (D)

Osmotic pressure (π) is directly proportional to the molar concentration of the solute and the number of ions produced when the solute dissociates.

• BaCl₂ dissociates into 3 ions (Ba²⁺, 2Cl⁻) per formula unit, so the effective concentration of ions is 0.01 × 3 = 0.03 M.
• NaCl dissociates into 2 ions (Na⁺, Cl⁻) per formula unit, so the effective concentration of ions is 0.005 × 2 = 0.01 M.

Therefore, the ratio of osmotic pressures is:

Thus, the correct answer is (A). 3:1

This is a buffer solution formed by benzoic acid (a weak acid) and sodium benzoate (its conjugate base). To calculate the pH of this buffer, we use the Henderson-Hasselbalch equation:

pH =

Where:

• pK_a = 4.2 (given)
• [A^-] is the concentration of the conjugate base (benzoate ion)
• [HA] is the concentration of the weak acid (benzoic acid)

First, calculate the moles of benzoic acid and sodium benzoate:

Moles of benzoic acid =

Moles of sodium benzoate =

The total volume of the solution is 1.0 L, so the concentrations are:

• [HA] =
• [A^-] =

Now, use the Henderson-Hasselbalch equation:

pH =

pH

pH = 4.4

Thus, the pH of the solution is 4.4.