pH of Weak Acids: Videos & Practice Problems

Weak acids are classified as weak electrolytes, partially ionizing in aqueous solutions, unlike strong acids which ionize completely. To determine the equilibrium concentrations of a weak acid, an ICE (Initial, Change, Equilibrium) chart is used, with molarity as the unit of measurement. The acid dissociation constant, Ka, is essential for these calculations. The percent ionization of a weak acid, which is less than 100%, indicates the extent of its ionization and is calculated using the formula:

$\left(\frac{\text{equilibrium concentration of}{H}_{3}O+}{\text{initial concentration of the weak acid}}\right)\times 100$

This calculation requires knowing or determining the hydronium ion concentration at equilibrium, often through an ICE chart. Understanding these concepts is crucial for analyzing acid behavior in solutions.

ICE Charts of Weak Acids

Weak acids are classified as weak electrolytes because they only partially dissociate into ions in solution. To analyze the equilibrium concentrations of these acids, we utilize an ICE chart, which stands for Initial, Change, and Equilibrium. This chart helps in organizing the concentrations of reactants and products at different stages of the reaction.

In the context of weak acids, the units used in the ICE chart are expressed in molarity (M). The acid dissociation constant, denoted as K_a, is crucial for these calculations as it quantifies the strength of the weak acid by measuring the extent of its dissociation. The equilibrium expression for a weak acid can be represented as:

$$ K_a = \frac{[H^+][A^-]}{[HA]} $$

In this equation, [H⁺] represents the concentration of hydrogen ions, [A^-] is the concentration of the conjugate base, and [HA] is the concentration of the undissociated weak acid at equilibrium. By applying the ICE chart, one can systematically determine the initial concentrations, the changes that occur as the reaction progresses, and the final equilibrium concentrations for any weak acid in question.

Hydronium Ion Concentration Example

To calculate the hydronium ion concentration for a 0.30 M solution of hydrocyanic acid (HCN), we first recognize that HCN is a weak acid, indicated by its acid dissociation constant (Ka) of \(4.9 \times 10^{-10}\), which is less than 1. This means it does not fully dissociate in solution.

We begin by setting up an ICE (Initial, Change, Equilibrium) table to analyze the dissociation of HCN in water. The dissociation reaction can be represented as:

\[ \text{HCN} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{CN}^- \]

In the ICE table, we denote the initial concentration of HCN as 0.30 M, while the initial concentrations of the products, hydronium ions (\(H_3O^+\)) and cyanide ions (\(CN^-\)), are both 0. The changes in concentration as the reaction reaches equilibrium are represented as follows:

Initial:
\[\text{[HCN]} = 0.30, \quad \text{[H}_3\text{O}^+] = 0, \quad \text{[CN}^-] = 0\]

Change:
\[\text{[HCN]} = -x, \quad \text{[H}_3\text{O}^+] = +x, \quad \text{[CN}^-] = +x\]

At equilibrium, the concentrations are:

Equilibrium:
\[\text{[HCN]} = 0.30 - x, \quad \text{[H}_3\text{O}^+] = x, \quad \text{[CN}^-] = x\]

Next, we can express the equilibrium constant expression for the dissociation of HCN:

\[ K_a = \frac{[\text{H}_3\text{O}^+][\text{CN}^-]}{[\text{HCN}]} \]

Substituting the equilibrium concentrations into the expression gives:

\[ 4.9 \times 10^{-10} = \frac{x^2}{0.30 - x} \]

To simplify the calculation, we can apply the 500 approximation method. This method allows us to ignore the \(-x\) in the denominator if the ratio of the initial concentration to \(K_a\) is greater than 500. Here, we calculate:

\[ \frac{0.30}{4.9 \times 10^{-10}} \approx 6.12 \times 10^8 \] (which is indeed greater than 500)

Thus, we can simplify our equation to:

\[ 4.9 \times 10^{-10} = \frac{x^2}{0.30} \]

Cross-multiplying gives:

\[ x^2 = 4.9 \times 10^{-10} \times 0.30 = 1.47 \times 10^{-10} \]

Taking the square root of both sides yields:

\[ x = \sqrt{1.47 \times 10^{-10}} \approx 1.2 \times 10^{-5} \, \text{M} \]

Since \(x\) represents the concentration of hydronium ions at equilibrium, we conclude that the hydronium ion concentration for the 0.30 M solution of hydrocyanic acid is:

\[ [H_3O^+] = 1.2 \times 10^{-5} \, \text{M} \]

Calculating pH of Weak Acids Example

To calculate the pH of a weak acid solution, such as a 0.074 M nitrous acid (HNO₂) solution, we first recognize that nitrous acid is a weak acid with a dissociation constant (K_a) of 4.6 × 10⁻⁴. Since K_a is less than 1, we can use an ICE (Initial, Change, Equilibrium) chart to determine the equilibrium concentrations of the species involved in the dissociation reaction:

HNO₂(aq) + H₂O(l) ⇌ NO⁻(aq) + H₃O⁺(aq)

In the ICE chart, we start with the initial concentration of the weak acid (0.074 M) and assume that the initial concentrations of the products (NO⁻ and H₃O⁺) are zero. The change in concentration for the reactants will be represented as -x, while the products will increase by +x:

Initial: 0.074, 0, 0

Change: -x, +x, +x

Equilibrium: 0.074 - x, x, x

Next, we set up the equilibrium expression based on the K_a value:

K_a = \(\frac{[NO^-][H_3O^+]}{[HNO_2]}\) = \(\frac{x^2}{0.074 - x}\)

To simplify the calculation, we can use the approximation method. We check if the initial concentration divided by K_a is greater than 500:

\(\frac{0.074}{4.6 \times 10^{-4}} \approx 160.87\)

Since this value is not greater than 500, we cannot ignore x in the denominator, and we must solve the quadratic equation:

4.6 × 10⁻⁴ (0.074 - x) = x²

Rearranging gives us:

x² + 4.6 × 10⁻⁴x - 3.404 × 10⁻⁵ = 0

Using the quadratic formula, where a = 1, b = 4.6 × 10⁻⁴, and c = -3.404 × 10⁻⁵:

x = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Calculating the discriminant:

x = \(\frac{-4.6 \times 10^{-4} \pm \sqrt{(4.6 \times 10^{-4})^2 - 4(1)(-3.404 \times 10^{-5})}}{2(1)}\)

Solving this yields two potential values for x. However, only the positive value is physically meaningful, as concentrations cannot be negative. The valid x value is approximately 0.005609 M, which represents the concentration of H₃O⁺.

Finally, we can calculate the pH using the formula:

pH = -log[H₃O⁺] = -log(0.005609) ≈ 2.25

Thus, the pH of the 0.074 M nitrous acid solution is approximately 2.25.

Calculating Percent Ionization of Weak Acids

Percent ionization, also known as percent dissociation, is a crucial concept in understanding the behavior of weak acids in aqueous solutions. It quantifies the extent to which a weak acid can ionize, reflecting its strength as an electrolyte. Unlike strong acids, which fully ionize (100%) in solution, weak acids only partially ionize, resulting in a percentage that is less than 100%.

To calculate the percent ionization of a weak acid, the following formula is used:

\[\text{Percent Ionization} = \left( \frac{[\text{H}_3\text{O}^+]}{[\text{HA}]_0} \right) \times 100\]

In this equation, \([\text{H}_3\text{O}^+]\) represents the equilibrium concentration of hydronium ions, while \([\text{HA}]_0\) denotes the initial concentration of the weak acid before any ionization occurs. To find the equilibrium concentration of hydronium ions, one can either be provided with this value or calculate it using an ICE (Initial, Change, Equilibrium) chart, which helps track the changes in concentration throughout the ionization process.

Understanding percent ionization is essential for predicting the behavior of weak acids in various chemical contexts, as it provides insight into their reactivity and strength in solution.

Percent Dissociation Example

To calculate the percent dissociation of acetic acid, we start with its initial concentration, which is given as 4.10 × 10⁻¹ M (or 0.410 M). Acetic acid is a weak acid, indicated by its acid dissociation constant (K_a) value of 1.8 × 10⁻⁵, which is less than 1. This means that it does not fully dissociate in solution.

We can set up an ICE (Initial, Change, Equilibrium) chart to analyze the dissociation of acetic acid (CH₃COOH) in water:

Species	Initial (M)	Change (M)	Equilibrium (M)
CH3COOH	0.410	−x	0.410 − x
H3O+	0	+x	x
CH3COO−	0	+x	x

Next, we can write the expression for the acid dissociation constant:

K_a = \(\frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}\)

Substituting the equilibrium concentrations into the expression gives us:

1.8 × 10⁻⁵ = \(\frac{x \cdot x}{0.410 - x}\)

To simplify the calculation, we can use the 500 approximation method. This method states that if the ratio of the initial concentration to K_a is greater than 500, we can ignore the term "−x" in the denominator. Here, we calculate:

\(\frac{0.410}{1.8 \times 10^{-5}} \approx 22,777.8\), which is indeed greater than 500.

Thus, we can simplify our equation to:

1.8 × 10⁻⁵ = \(\frac{x^2}{0.410}\)

Cross-multiplying gives:

x² = 1.8 × 10⁻⁵ × 0.410 = 7.38 × 10⁻⁶

Taking the square root of both sides, we find:

x = 2.717 × 10⁻³ M

This value of x represents the concentration of H₃O⁺ at equilibrium. To find the percent dissociation, we use the formula:

Percent Dissociation = \(\frac{[H_3O^+]}{[CH_3COOH]_{initial}} \times 100\)

Substituting the values we have:

Percent Dissociation = \(\frac{2.717 \times 10^{-3}}{0.410} \times 100 \approx 0.66\%\)

This indicates that approximately 0.66% of the acetic acid dissociates in solution, which is consistent with the behavior of weak acids.

Calculate the [H⁺] of a 0.50 M solution of methylammonium bromide, CH₃NH₃Br. The K_b of methylamine, CH₃NH₂, is given as 4.4 x 10^-4.

• A. 0.50 M
• B. 1.14 × 10⁻¹¹ M
• C. 3.4 × 10⁻⁶ M
• D. 2.97 × 10⁻⁹ M

Answer: (C)

An unknown weak acid has an initial concentration of 0.55 M. What is the pH of the solution if the weak acid also has a pK_a of 5.79?

• A. 0.60
• B. 6.05
• C. 3.02
• D. 5.75

Answer: (C)

Here’s what students ask on this topic:

What are the assumptions when calculating the pH of weak acids?

When calculating the pH of weak acids, several assumptions are made to simplify the process:

1. The acid is only partially ionized in solution, meaning it does not dissociate completely as strong acids do.
2. The concentration of the acid at equilibrium is approximately equal to the initial concentration since the degree of ionization is low.
3. The change in concentration of the acid during ionization is so small that it can be ignored. This is because the amount of acid that dissociates is much less than the amount that remains undissociated.
4. The concentration of hydronium ions ($H_{3}O+$) produced by the ionization of the acid is equal to the concentration of the conjugate base formed.
5. The contribution of hydronium ions from the autoionization of water (2H2O⇌H3O++OH−) is negligible compared to the concentration of hydronium ions from the weak acid.

These assumptions allow the use of the simplified version of the acid dissociation constant (Ka) expression to solve for pH, which is practical for weak acids with a Ka much less than 1.

How do you calculate the pH of weak acids?

To calculate the pH of a weak acid, you'll need to use the acid dissociation constant (Ka) for the acid in question. Here's a step-by-step guide:

1. Write down the equilibrium expression for the dissociation of the weak acid (HA):
   $\mathrm{HA}\rightleftharpoons H^{+}+A^{-}$
2. Write the expression for the acid dissociation constant (Ka):
   $Ka=\frac{{\mathrm{[H]}}^{+}{\mathrm{[A]}}^{-}}{\mathrm{[HA]}}$
3. Set up an ICE table (Initial, Change, Equilibrium) to determine the concentrations of the species at equilibrium. Assume the initial concentration of H+ and A- is zero and the concentration of HA is the initial concentration of the acid.
4. Since the acid is weak, assume that the change in concentration of HA is small and can be approximated as x, where x is the concentration of H+ and A- at equilibrium.
5. Plug the equilibrium concentrations back into the Ka expression and solve for x, which is [H+].
6. Calculate the pH by taking the negative logarithm (base 10) of the H+ concentration:
   $\mathrm{pH}=-\log \left[H^{+}\right]$

Remember that if the percent ionization is greater than 5%, the assumption that the change in concentration of HA is small may not be valid, and a more accurate method may be needed.