pH Revisited: Videos & Practice Problems

pH is calculated as the negative logarithm of the hydronium ion concentration (H₃O⁺). In pure water at 50°C, the ion product constant (K_w) is 7.94 × 10⁻¹⁴, leading to a pH of approximately 6.55. When free ions are present, ionic strength and activity coefficients must be considered, affecting the pH calculation. For example, in a saturated barium hydroxide solution with lithium nitrite, the pH is influenced by the ionic strength, resulting in a pH of about 13.41 after accounting for activity coefficients.

Glass Electrodes

Understanding pH is essential in chemistry, as it represents the negative logarithm of the hydrogen ion concentration, expressed as pH = -\log[H^+] or pH = -\log[H_3O^+]. At elevated temperatures, such as 50 degrees Celsius, the ionization constant of water, denoted as K_w, changes. For pure water at this temperature, K_w is given as 7.94 \times 10^{-14}.

In a neutral solution, the concentrations of hydrogen ions [H^+] and hydroxide ions [OH^-] are equal. We can denote both concentrations as x. The relationship between these ions is defined by the equation:

[H^+][OH^-] = K_w

Substituting x for both concentrations gives us:

x^2 = 7.94 \times 10^{-14}

To find x, we take the square root of both sides:

x = \sqrt{7.94 \times 10^{-14}} \approx 2.8178 \times 10^{-7} \, \text{M}

This value represents both the hydrogen ion concentration and the hydroxide ion concentration in the neutral solution. To calculate the pH, we focus on the hydrogen ion concentration:

pH = -\log(2.8178 \times 10^{-7}) \approx 6.55

This result indicates that the pH of pure water at 50 degrees Celsius is approximately 6.55, which is slightly acidic compared to the neutral pH of 7 at 25 degrees Celsius.

When considering pure water with additional free-floating ions, it is important to account for the ionic strength and the activity coefficients, as these factors can influence the pH. The presence of ions alters the behavior of the solution, affecting both the concentration of hydrogen ions and the overall pH. Understanding these concepts is crucial for accurately determining pH in various chemical contexts.

Glass Electrodes

In this discussion, we explore the relationship between ionic strength and the activity coefficients of ions in a solution, particularly focusing on the behavior of hydrogen ions (H⁺) and hydroxide ions (OH^-) in the presence of non-common ions like sodium (Na⁺) and bromide (Br^-). When dealing with pure water, the concentrations of H⁺ and OH^- are equal, denoted as x. However, the introduction of other ions affects the ionic strength of the solution, which in turn influences the activity coefficients of H⁺ and OH^-.

The ionic strength (I) can be calculated using the formula:

I = 0.5 * (C_Na+ * z_Na+² + C_Br- * z_Br-²)

where C represents the concentration of each ion and z represents the charge of each ion. For sodium bromide, the ionic strength is determined to be 0.05. This value is then used to look up the activity coefficients for H⁺ and OH^- in an activity coefficients table, yielding values of 0.933 and 0.926, respectively.

At a temperature of 50 degrees Celsius, the ion product of water (K_w) is 7.94 x 10^-14. The relationship between the concentrations and activity coefficients can be expressed as:

[H⁺][OH^-] = K_w

Substituting the activity coefficients into the equation gives:

(0.933 * x)(0.926 * x) = 7.94 x 10^-14

Solving for x involves multiplying the activity coefficients and rearranging the equation to find x² = 9.19 x 10^-14, leading to x = 3.03 x 10^-7 M for both H⁺ and OH^- concentrations.

To find the pH, we use the formula:

pH = -log([H⁺])

However, we must account for the activity coefficient of H⁺:

pH = -log(3.03 x 10^-7 * 0.933)

This calculation results in a pH of approximately 6.54, slightly different from the initial pH of 6.55, demonstrating the impact of ionic strength on pH calculations. If the concentration of ions or their charges were increased, the effect on pH would be more pronounced.

In a similar context, when analyzing the solubility product constant (K_sp) of barium hydroxide in the presence of lithium nitrate, the same principles apply. The presence of non-common ions necessitates consideration of ionic strength and activity coefficients, reinforcing the importance of these factors in accurately determining ion concentrations and pH in solutions.

Glass Electrodes

To determine the pH of a saturated solution of barium hydroxide (\( \text{Ba(OH)}_2 \)) dissolved in a 0.05 M lithium nitrite (\( \text{LiNO}_2 \)) solution, we start by recognizing the solubility product constant (\( K_{sp} \)) of barium hydroxide, which is \( 5.0 \times 10^{-3} \). When barium hydroxide dissolves, it dissociates into barium ions (\( \text{Ba}^{2+} \)) and hydroxide ions (\( \text{OH}^- \)) according to the equation:

\[ \text{Ba(OH)}_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2 \text{OH}^- (aq) \]

In this case, we denote the concentration of barium ions as \( x \) and the concentration of hydroxide ions as \( 2x \). The expression for \( K_{sp} \) can be written as:

\[ K_{sp} = [\text{Ba}^{2+}] \cdot [\text{OH}^-]^2 \]

Since we are dealing with a solution that contains non-common ions, the ionic strength must be considered, which affects the activity coefficients of the ions. The ionic strength (\( I \)) is calculated using the concentrations of lithium ions (\( \text{Li}^+ \)) and nitrate ions (\( \text{NO}_3^- \)), both of which contribute equally due to their charges:

\[ I = 0.05 \times 1^2 + 0.05 \times 1^2 = 0.1 \]

Next, we look up the activity coefficients for \( \text{Ba}^{2+} \) and \( \text{OH}^- \) ions, which are found to be approximately 0.465 and 0.81, respectively. Incorporating these coefficients into the \( K_{sp} \) expression gives us:

\[ K_{sp} = (0.465 \cdot x) \cdot (0.81 \cdot 2x)^2 \]

Expanding this, we have:

\[ K_{sp} = 0.465 \cdot x \cdot (0.81^2 \cdot 4x^2) = 0.465 \cdot 0.6561 \cdot 4x^3 \approx 1.22035 \cdot x^3 \]

Setting this equal to the \( K_{sp} \) value, we find:

\[ 5.0 \times 10^{-3} = 1.22035 \cdot x^3 \]

Solving for \( x \), we get:

\[ x^3 = \frac{5.0 \times 10^{-3}}{1.22035} \approx 0.004097 \]

Taking the cube root yields:

\[ x \approx 0.16 \, \text{M} \]

Since the concentration of hydroxide ions is \( 2x \), we calculate:

\[ [\text{OH}^-] = 2 \cdot 0.16 \approx 0.32 \, \text{M} \]

Next, we find the activity of hydroxide ions by multiplying the concentration by its activity coefficient:

\[ a_{\text{OH}^-} = [\text{OH}^-] \cdot \text{activity coefficient} = 0.32 \cdot 0.81 \approx 0.2592 \]

To find the pOH, we take the negative logarithm of the hydroxide activity:

\[ \text{pOH} = -\log(0.2592) \approx 0.586 \]

Finally, we can calculate the pH using the relationship:

\[ \text{pH} = 14 - \text{pOH} \approx 14 - 0.586 \approx 13.41 \]

This calculation illustrates the impact of ionic strength and activity coefficients on the solubility and pH of ionic compounds in solution. The presence of non-common ions enhances the solubility of barium hydroxide, leading to a higher concentration of hydroxide ions and, consequently, a higher pH.

Here’s what students ask on this topic:

What is the pH of pure water at 50°C?

The pH of pure water at 50°C can be calculated using the ion product constant (Kw) for water at this temperature, which is 7.94 × 10⁻¹⁴. In pure water, the concentration of H⁺ and OH⁻ ions are equal. Therefore, we set [H⁺] = [OH⁻] = x. Since Kw = [H⁺][OH⁻], we have x² = 7.94 × 10⁻¹⁴. Solving for x, we get x = √(7.94 × 10⁻¹⁴) = 2.8178 × 10⁻⁷ M. The pH is then calculated as pH = −log[H⁺] = −log(2.8178 × 10⁻⁷) ≈ 6.55.

How do free floating ions affect the pH of a solution?

Free floating ions in a solution affect the pH by influencing the ionic strength and activity coefficients of the ions involved. When non-common ions are present, they increase the ionic strength of the solution, which in turn affects the activity coefficients of H⁺ and OH⁻ ions. This means that the effective concentration of these ions is altered, leading to a different pH value. For example, in a solution with Na⁺ and Br⁻ ions, the activity coefficients must be considered to accurately calculate the pH.

How do you calculate the pH of a solution with non-common ions?

To calculate the pH of a solution with non-common ions, you need to account for the ionic strength and activity coefficients. First, determine the ionic strength (I) using the formula I = 0.5 Σc_iz_i², where c_i is the concentration and z_i is the charge of each ion. Then, use the ionic strength to find the activity coefficients (γ) from a table. The effective concentration of H⁺ is then [H⁺] = γ[H⁺]. Finally, calculate the pH as pH = −log([H⁺]γ).

What is the pH of a saturated solution of barium hydroxide in 0.05 M lithium nitrite?

To find the pH of a saturated solution of barium hydroxide in 0.05 M lithium nitrite, first calculate the ionic strength (I) of the solution. For lithium nitrite, I = 0.05. Using the activity coefficients for Ba2+ and OH− from a table, we find γBa2+ = 0.465 and γOH− = 0.81. The Ksp of Ba(OH)2 is 5.0 × 10−3. Set up the equation Ksp = [Ba2+][OH−]2γBa2+γOH−2. Solving for [OH−], we get [OH−] = 0.032 M. The activity of OH− is [OH−]γOH− = 0.032 × 0.81 = 0.0259. The pOH is −log(0.0259) ≈ 0.586, and the pH is 14 − 0.586 ≈ 13.41.

Why is it important to consider activity coefficients in pH calculations?

Activity coefficients are important in pH calculations because they account for the non-ideal behavior of ions in solution. In solutions with high ionic strength, interactions between ions affect their effective concentration, which is different from their actual concentration. The activity coefficient (γ) corrects for these interactions, providing a more accurate measure of the ion's activity. Ignoring activity coefficients can lead to significant errors in pH calculations, especially in solutions with high ionic strength or non-common ions.