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13) What is the pH of a 0.001 M solution of Ba(OH)2? a) 11.00 b) 3.00 с). 11.13 d) 2.87

13) What is the pH of a 0.001 M solution of Ba(OH)2? a) 11.00 b) 3.00 с). 11.13 d) 2.87

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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**Question 13:** What is the pH of a 0.001 M solution of Ba(OH)₂? **Options:** - a) 11.00 - b) 3.00 - c) 11.13 - d) 2.87 **Explanation:** To determine the pH of a 0.001 M solution of Ba(OH)₂, we first recognize that Ba(OH)₂ is a strong base and dissociates completely in water to produce barium ions (Ba²⁺) and hydroxide ions (OH⁻). The concentration of OH⁻ ions will be twice the concentration of Ba(OH)₂ because each formula unit of Ba(OH)₂ produces two OH⁻ ions. Therefore, the concentration of OH⁻ ions is 0.002 M. To find the pOH, use the formula: \[ \text{pOH} = -\log[\text{OH}^-] \] \[ \text{pOH} = -\log(0.002) = 2.7 \] Finally, to find the pH, use the relationship: \[ \text{pH} + \text{pOH} = 14 \] \[ \text{pH} = 14 - 2.7 = 11.3 \] Matching this with the options provided, choice c) 11.13 is the closest approximation.
Transcribed Image Text:**Question 13:** What is the pH of a 0.001 M solution of Ba(OH)₂? **Options:** - a) 11.00 - b) 3.00 - c) 11.13 - d) 2.87 **Explanation:** To determine the pH of a 0.001 M solution of Ba(OH)₂, we first recognize that Ba(OH)₂ is a strong base and dissociates completely in water to produce barium ions (Ba²⁺) and hydroxide ions (OH⁻). The concentration of OH⁻ ions will be twice the concentration of Ba(OH)₂ because each formula unit of Ba(OH)₂ produces two OH⁻ ions. Therefore, the concentration of OH⁻ ions is 0.002 M. To find the pOH, use the formula: \[ \text{pOH} = -\log[\text{OH}^-] \] \[ \text{pOH} = -\log(0.002) = 2.7 \] Finally, to find the pH, use the relationship: \[ \text{pH} + \text{pOH} = 14 \] \[ \text{pH} = 14 - 2.7 = 11.3 \] Matching this with the options provided, choice c) 11.13 is the closest approximation.
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