Ch.16 - Acid-Base Equilibria

Brown - Chemistry: The Central Science 15th Edition

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Brown 15th Edition

Ch.16 - Acid-Base Equilibria

Problema 84b

Capitolo 16, Problema 84b

Using data from Appendix D, calculate 3OH-4 and pH for each of the following solutions: (b) 0.035 M Na2S

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Identify the relevant chemical species in the solution. Sodium sulfide (Na2S) dissociates in water to form 2 Na^+ ions and 1 S^2- ion.

Recognize that the sulfide ion (S^2-) can react with water in a hydrolysis reaction to form hydroxide ions (OH^-). The reaction is: S^{2-} + H_2O \(\rightleftharpoons\) HS^- + OH^-.

Write the equilibrium expression for the hydrolysis reaction. The equilibrium constant for this reaction is the base dissociation constant, K_b, which can be found using the relationship K_w = K_a \(\times\) K_b, where K_w is the ion-product constant of water and K_a is the acid dissociation constant for the conjugate acid (HS^-).

Calculate the concentration of OH^- ions produced by the hydrolysis of S^2-. Use the initial concentration of S^2- (0.035 M) and the equilibrium expression to solve for [OH^-].

Determine the pH of the solution. First, calculate the pOH using the concentration of OH^- ions: \(\text{pOH}\) = -\(\log\)[OH^-]. Then, use the relationship \(\text{pH}\) + \(\text{pOH}\) = 14 to find the pH.

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Dissociation of Sodium Sulfide (Na2S)

Sodium sulfide (Na2S) dissociates in water to produce sodium ions (Na+) and sulfide ions (S2-). The sulfide ion can further react with water to form hydroxide ions (OH-) and hydrogen sulfide (H2S), leading to an increase in the concentration of OH- in the solution. Understanding this dissociation is crucial for calculating the concentration of hydroxide ions in the solution.

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04:52

Percent Dissociation Example

Hydroxide Ion Concentration (OH-)

The concentration of hydroxide ions (OH-) is a key factor in determining the basicity of a solution. In this case, the dissociation of Na2S produces OH- ions, which can be calculated based on the stoichiometry of the reaction. The concentration of OH- is essential for calculating the pH of the solution, as it directly influences the acidity or basicity.

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06:12

Hydroxide Ion Concentration Example

pH Calculation

pH is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydrogen ion concentration (H+). For basic solutions, pH can be calculated using the relationship between pH and pOH, where pH + pOH = 14. By first determining the concentration of OH- from the dissociation of Na2S, one can find pOH and subsequently calculate the pH of the solution.

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02:15

pH Calculation Example

Pratica correlata

Domanda del libro di testo

Pyridinium bromide 1C5H5NHBr2 is a strong electrolyte that dissociates completely into C5H5NH+ and Br-. An aqueous solution of pyridinium bromide has a pH of 2.95. (b) Using Appendix D, calculate the Ka for pyridinium bromide.

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Domanda del libro di testo

Given that K_b for ammonia is 1.8 × 10^-5 and that for hydroxylamine is 1.1 × 10^-8, which is the stronger base?

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Domanda del libro di testo

Which is the stronger acid, the ammonium ion or the hydroxylammonium ion?

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Domanda del libro di testo

Pyridinium bromide 1C5H5NHBr2 is a strong electrolyte that dissociates completely into C5H5NH+ and Br-. An aqueous solution of pyridinium bromide has a pH of 2.95. (a) Write out the reaction that leads to this acidic pH.

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