Glass Electrodes and pH Measurements: 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 concentrations of H⁺ and OH^- 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) = 2.82 × 10^-7 M. The pH is then calculated as pH = -log[H⁺] = -log(2.82 × 10^-7) ≈ 6.55.

How do ionic strength and activity coefficients affect pH measurements?

Ionic strength and activity coefficients significantly impact pH measurements, especially in solutions with free-floating ions. Ionic strength (μ) is a measure of the total concentration of ions in a solution and affects the activity coefficients (γ) of ions. The activity of an ion is given by its concentration multiplied by its activity coefficient. For accurate pH calculations, the activity of H⁺ ions must be considered: pH = -log(a_{H⁺) = -log([H⁺]γH⁺). Higher ionic strength generally decreases the activity coefficients, leading to a lower pH than calculated using concentrations alone.}

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 must account for ionic strength and activity coefficients. First, determine the ionic strength (μ) using the formula: μ = 0.5 Σc_iz_i², where c_i is the concentration and z_i is the charge of each ion. Next, use tables to find the activity coefficients (γ) for H⁺ and OH^- at the calculated ionic strength. Finally, calculate the pH using the adjusted concentration: pH = -log([H⁺]γ_H⁺).

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

To find the pH of a saturated barium hydroxide solution in 0.05 M lithium nitrite, first calculate the ionic strength (μ) of the solution. For lithium nitrite, μ = 0.05. Using activity coefficient tables, find γ_Ba²⁺ and γ_OH^-. For Ba(OH)₂, K_sp = [Ba²⁺][OH^-]². Let [Ba²⁺] = x and [OH^-] = 2x. Adjust for activity coefficients: K_sp = (xγ_Ba²⁺)(2xγ_OH^-)². Solve for x, then find [OH^-]. Calculate pOH = -log([OH^-]γ_OH^-), and pH = 14 - pOH. The pH is approximately 13.41.

Why is the pH of pure water not always 7?

The pH of pure water is not always 7 because the ion product constant (Kw) of water is temperature-dependent. At 25°C, Kw is 1.0 × 10^-14, leading to a pH of 7. However, at different temperatures, Kw changes. For example, at 50°C, Kw is 7.94 × 10^-14, resulting in a pH of approximately 6.55. This variation occurs because the degree of water ionization changes with temperature, affecting the concentrations of H⁺ and OH^- ions.