Weak Acid-Base Equilibria: Videos & Practice Problems

To determine the original molarity of a weak acid with a given acid ionization constant (Ka) of 4.7 × 10^-3 and a pH of 4.12, first calculate the hydronium ion concentration using the formula ${10}^{-}$, resulting in 7.59 × 10^-5 M. Set up the equilibrium expression $K=\frac{\mathrm{[A^-]}}{\mathrm{[H₃O⁺]}}$ and solve for the formal concentration, yielding 7.71 × 10^-5 M.

Weak Acid-Base Equilibria Calculations 1

To determine the original molarity of a weak acid solution with a given acid dissociation constant (Ka) and pH, we start by recognizing that a weak acid partially dissociates in water. The provided Ka value of \(4.7 \times 10^{-3}\) indicates that it is indeed a weak acid, as its value is less than 1.

Given the pH of the solution is 4.12, we can calculate the concentration of hydronium ions (\(H_3O^+\)) using the formula:

\( [H_3O^+] = 10^{-\text{pH}} \)

Substituting the pH value, we find:

\( [H_3O^+] = 10^{-4.12} \approx 7.59 \times 10^{-5} \, \text{M} \)

In the dissociation of the weak acid (HA), the equilibrium can be represented as:

HA ⇌ A^- + H₃O⁺

Using an ICE (Initial, Change, Equilibrium) table, we denote the initial concentration of the weak acid as \(f\), and at equilibrium, the concentrations are:

• HA: \(f - x\)
• A^-: \(x\)
• H₃O⁺: \(x\)

Since both products are equal to \(x\), we can express the equilibrium constant expression as:

\(K_a = \frac{[A^-][H_3O^+]}{[HA]} = \frac{x^2}{f - x}\)

Substituting the known values, we have:

\(4.7 \times 10^{-3} = \frac{(7.59 \times 10^{-5})^2}{f - 7.59 \times 10^{-5}}\)

Cross-multiplying and simplifying leads to:

\(4.7 \times 10^{-3} (f - 7.59 \times 10^{-5}) = (7.59 \times 10^{-5})^2\)

Calculating the square of \(7.59 \times 10^{-5}\) gives approximately \(5.75 \times 10^{-9}\). Expanding the left side results in:

\(4.7 \times 10^{-3} f - 3.56 \times 10^{-7} = 5.75 \times 10^{-9}\)

To isolate \(f\), we rearrange the equation:

\(4.7 \times 10^{-3} f = 5.75 \times 10^{-9} + 3.56 \times 10^{-7}\)

Calculating the right side yields approximately \(3.62 \times 10^{-7}\). Finally, we solve for \(f\):

\(f = \frac{3.62 \times 10^{-7}}{4.7 \times 10^{-3}} \approx 7.71 \times 10^{-5} \, \text{M}\)

This value represents the original molarity of the weak acid solution. Key concepts to remember include the relationship between pH and \(H_3O^+\) concentration, the significance of Ka for weak acids, and the use of equilibrium expressions to solve for unknown concentrations.

Weak Acid-Base Equilibria Calculations 1

To identify an unknown monoprotic acid, we can determine its acid dissociation constant, \( K_a \), using the concentration of the acid solution and its pH. In this scenario, a solution with a concentration of \( 6.05 \times 10^{-2} \) M has a pH of 2.122. Since we are dealing with a weak acid, we recognize that strong acids have \( K_a \) values that approach infinity, indicating complete dissociation.

The dissociation of a weak acid can be represented as follows:

\[ HA \rightleftharpoons H^+ + A^- \]

In this reaction, \( HA \) is the weak acid, \( H^+ \) is the hydrogen ion, and \( A^- \) is the conjugate base. To analyze the equilibrium, we can set up an ICE (Initial, Change, Equilibrium) table:

ICE Table:

Species	Initial (M)	Change (M)	Equilibrium (M)
HA	6.05 × 10-2	-x	6.05 × 10-2 - x
H+	0	+x	x
A-	0	+x	x

At equilibrium, the expression for \( K_a \) is given by:

\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{x^2}{6.05 \times 10^{-2} - x} \]

To find \( x \), we can calculate the concentration of \( H^+ \) ions using the pH:

\[ [H^+] = 10^{-\text{pH}} = 10^{-2.122} \approx 7.55 \times 10^{-3} \, \text{M} \]

Thus, \( x = 7.55 \times 10^{-3} \). Substituting this value into the \( K_a \) expression gives:

\[ K_a = \frac{(7.55 \times 10^{-3})^2}{6.05 \times 10^{-2} - 7.55 \times 10^{-3}} \]

Calculating this yields:

\[ K_a \approx 1.08 \times 10^{-3} \]

It is important to note that equilibrium constants like \( K_a \) are dimensionless and do not have units. This value indicates the strength of the weak acid, with lower \( K_a \) values corresponding to weaker acids. In this case, the calculated \( K_a \) value confirms the identity of the unknown monoprotic acid.

Weak Acid-Base Equilibria Calculations

To determine the hydronium ion concentration in a 0.10 molar solution of a weak acid with a given acid dissociation constant (Ka) of 5.35, we first need to convert pKa to Ka. The relationship is defined as:

$$K_a = 10^{-\text{pKa}}$$

Substituting the provided pKa value:

$$K_a = 10^{-5.35} \approx 4.467 \times 10^{-6}$$

Next, we set up an ICE (Initial, Change, Equilibrium) chart to analyze the dissociation of the weak acid (HA) in water:

$$\text{HA} \rightleftharpoons \text{A}^- + \text{H}_3\text{O}^+$$

In the ICE chart, the initial concentration of HA is 0.10 M, while the initial concentrations of A^- and H₃O⁺ are both 0. At equilibrium, we denote the change in concentration of H₃O⁺ and A^- as x:

Initial:      [HA] = 0.10      [A^-] = 0      [H₃O⁺] = 0

Change:      [HA] = -x      [A^-] = +x      [H₃O⁺] = +x

Equilibrium:      [HA] = 0.10 - x      [A^-] = x      [H₃O⁺] = x

Using the expression for Ka:

$$K_a = \frac{[\text{A}^-][\text{H}_3\text{O}^+]}{[\text{HA}]} = \frac{x \cdot x}{0.10 - x} = \frac{x^2}{0.10 - x}$$

Since we are dealing with a weak acid, we can apply the 5% approximation method. This method allows us to ignore x in the denominator if:

$$\frac{[\text{HA}]_0}{K_a} > 500$$

Calculating this gives:

$$\frac{0.10}{4.467 \times 10^{-6}} \approx 22,387.2$$

Since this value is greater than 500, we can simplify our equation to:

$$K_a \approx \frac{x^2}{0.10}$$

Substituting the value of Ka:

$$4.467 \times 10^{-6} = \frac{x^2}{0.10}$$

Multiplying both sides by 0.10:

$$4.467 \times 10^{-7} = x^2$$

Taking the square root of both sides gives:

$$x \approx 6.68 \times 10^{-4} \text{ M}$$

This value of x represents the concentration of hydronium ions (H₃O⁺) in the solution. Therefore, the hydronium ion concentration in the 0.10 molar solution of the weak acid is approximately 6.68 x 10^-4 M.

The pH of an aqueous 0.10 M nitrite ion is 8.17. What is the base dissociation constant of the base?

• A. 4.6 x 10^-16
• B. 2.2 x 10^-11
• C. 1.6 x 10^-6
• D. 1.6 x 10^‑5
• E. 1.2 x 10^-3

Answer: (B)

Here’s what students ask on this topic:

How do you determine the original molarity of a weak acid given its Ka and pH?

To determine the original molarity of a weak acid with a given Ka and pH, follow these steps: First, calculate the hydronium ion concentration using the formula ${10}^{-\mathrm{pH}}$. For example, if the pH is 4.12, the hydronium ion concentration is ${10}^{-4.12}$, which equals 7.59 × 10^-5 M. Next, set up the equilibrium expression $K=\frac{[A-]\left[H_{3}O\right]}{\left[}$ and solve for the formal concentration. This will yield the original molarity of the weak acid.

What is the significance of the Ka value in weak acid-base equilibria?

The Ka value, or acid ionization constant, is crucial in weak acid-base equilibria as it measures the strength of a weak acid. A smaller Ka value (less than 1) indicates a weaker acid, meaning it partially ionizes in solution. The Ka value helps predict the extent of ionization and the concentrations of the acid and its conjugate base at equilibrium. It is used in equilibrium expressions to calculate pH, pOH, and the concentrations of various species in the solution.

How do you use an ICE chart to solve weak acid equilibrium problems?

An ICE chart (Initial, Change, Equilibrium) is a systematic way to solve weak acid equilibrium problems. Start by listing the initial concentrations of the reactants and products. Then, indicate the changes in concentration as the system reaches equilibrium, using variables like x. Finally, express the equilibrium concentrations in terms of these variables. Use the equilibrium expression (Ka) to set up an equation and solve for the unknowns. This method helps organize the data and simplifies the calculation process.

How do you calculate the pH of a weak acid solution?

To calculate the pH of a weak acid solution, follow these steps: First, write the ionization equation of the weak acid. Use the given Ka value and set up the equilibrium expression. Create an ICE chart to determine the equilibrium concentrations of the species involved. Solve for the hydronium ion concentration [H₃O⁺]. Finally, calculate the pH using the formula $-{10}^{-\mathrm{pH}}$. For example, if [H₃O⁺] is 7.59 × 10^-5 M, the pH is -log(7.59 × 10^-5), which equals 4.12.

What is the relationship between pH and pOH in weak acid-base equilibria?

The relationship between pH and pOH in weak acid-base equilibria is given by the equation pH + pOH = 14 at 25°C. This relationship helps in determining the concentration of hydroxide ions (OH^-) if the pH is known, and vice versa. For example, if the pH of a solution is 4.12, the pOH can be calculated as 14 - 4.12 = 9.88. This relationship is essential for understanding the acid-base balance in a solution and for calculating the concentrations of H₃O⁺ and OH^- ions.