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Determining [OH-] and pH of a 0.246 M Solution of Ca(OH)?

Determining [OH-] and pH of a 0.246 M Solution of Ca(OH)?

February 21, 2025 Education

Determining [OH-] and pH of a 0.246 M Solution of Ca(OH)?

The determination of [OH-] and pH of a calcium hydroxide (Ca(OH)?) solution involves understanding the dissociation of Ca(OH)? in aqueous solution, the relationship between [OH-], pOH, and pH, and the solubility product (Ksp) of calcium hydroxide. This article elucidates the process and provides a step-by-step guide.

Basic Dissociation and pH Calculation

Calcium hydroxide (Ca(OH)?) dissociates into calcium ions (Ca2?) and hydroxide ions (OH?) in an aqueous solution. This process can be represented by the following dissociation equation:

Ca(OH)? larr;→ Ca2?   2OH?

To determine the pH of a 0.246 M solution of Ca(OH)?, we start by calculating [OH?] and then use it to find pH. Given the concentration, the calculation is as follows:

[OH?] 0.246 M

The pH can be calculated using the pH formula and the relationship between pH and pOH.

pH  -log[H?]pOH  -log[OH?]pOH  14 - pH

Given the concentration of OH?:

pOH  -log[0.246]  0.61pH  14 - 0.61  13.39

Therefore, the [OH?] and pH for the 0.246 M solution of Ca(OH)? are [OH?] 0.246 M and pH 13.39, respectively.

Calculation Using Dissociation Equilibrium

To get a more accurate estimation, we can use the dissociation equilibrium expression. Let x be the amount dissociated from Ca(OH)?:

Ca(OH)? (aq) ? Ca2? (aq)   2OH? (aq)

At equilibrium, the concentrations are:

[Ca2?]  x[OH?]  2x[Ca(OH)?]  0.246 - x

The dissociation constant (Kd) can be expressed as:

Kd  [Ca2?][OH?]^2 / [Ca(OH)?]

Assuming x is small compared to 0.246, Kd can be simplified to:

(x)(2x)^2  4x^3 / 0.246

Solving for x, we find:

x  0.0097 M[OH?]  2x  0.0194 M

Subsequently, using the relationship between pH and pOH, we calculate:

pH  14 - log[OH?]pH  14 - log(0.0194)pH  12.72

Thus, the [OH?] and pH for the 0.246 M solution of Ca(OH)? are [OH?] 0.0194 M and pH 12.72.

Considering the Solubility Product of Calcium Hydroxide

Calcium hydroxide is sparingly soluble in aqueous solution. The actual solubility is limited by its solubility product (Ksp). The Ksp expression for Ca(OH)? is:

Ksp  [Ca2?][OH?]^2

Given the Ksp value for calcium hydroxide (5.5 × 10??), we can calculate the molar solubility as follows:

Ksp  S(2S)2  4S35.5 × 10??  4S3S  (5.5 × 10??)^(1/3)  0.0173 M[OH?]  2S  0.0346 M

The pOH and pH of the solution are calculated as:

pOH  -log[OH?]pOH  -log(0.0346)  1.46pH  14 - pOH  12.54

Thus, considering the solubility product, the [OH?] and pH for the 0.246 M solution of Ca(OH)? are [OH?] 0.0346 M and pH 12.54.

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