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Science Chemistry Chemistry: An Atoms-Focused Approach

To estimate: The p H of 0.5 M solution of sodium salt of the weak acid using the graph given in Figure P15.3 of the textbook. Solution: The p H of 0.5 M solution of sodium salt of the weak acid is 9.75 . Explanation: Concept: From the plot in figure 15.3, we can see half equivalence point is at p H 5.8 . We know that at half equivalence point p H = p K a = 5.8 . From the p K a value, we can calculate K b value for conjugate base of weak acid and then [ O H - ] of the salt solution using ICE table and K b e q u a t i o n . Once we know the [ O H - ] concentration we can find the p H of the solution. Formula: i) p K a + p K b = 14 ii) p K b = - log K b iii) K b = O H - [ H A ] [ A - ] iv) p O H = - log O H - v) p H + p O H = 14 Given: Initial concentration of sodium salt of weak acid = 0.5 M Figure 15.3 of the textbook. Calculations: Let us assume that the sodium salt of the weak acid is N a A . The hydrolysis reaction of conjugate base A - of weak acid H A is written as: A - a q + H 2 O l ⇌ H A a q + O H - a q Calculating the K b value from the p K a = 5.8 . p K a + p K b = 14 p K b = 14 - p K a p K b = 14 - 5.8 = 8.2 Now, p k b = - log K b K b = 10 - p K b = 10 - 8.2 = 6.31 × 10 - 9 The concentration of the salt N a A is 0.5 M . Thus, the concentration of A - will also be 0.5 M . To determine [ O H - ] we have to draw ICE table for the above reaction R e a c t i o n A - a q + H 2 O l ⇌ H A a q + O H - a q A - M H A M O H - M I n i t i a l 0.5 0 0 C h a n g e - x + x + x E q u i l i b r i u m 0.5 - x x x K b = O H - H A A - 6.30 × 10 - 9 = x 2 ( 0.5 - x ) Because K b is small relative to initial concentration of A - , we can make the simplifying assumption that x is small compared with 0.5 M , and so 0.5 - x ≈ 0.5 6.30 × 10 - 9 = x 2 ( 0.5 ) x 2 = 6.30 × 10 - 9 × 0.5 = 3.15 × 10 - 9 x = O H - = 5.6167 × 10 - 5 M Now calculating p H of the solution from [ O H - ] p O H = - log O H - p O H = - log ⁡ 5.6167 × 10 - 5 M p O H = 4.25 p H + p O H = 14 p H = 14 - 4.25 = 9.75 Therefore, the p H of 0.5 M solution of sodium salt of the weak acid is 9.75 . Conclusion: At half equivalence point, the p H of the solution is equal to the p K a of the weak acid.

To estimate: The p H of 0.5 M solution of sodium salt of the weak acid using the graph given in Figure P15.3 of the textbook. Solution: The p H of 0.5 M solution of sodium salt of the weak acid is 9.75 . Explanation: Concept: From the plot in figure 15.3, we can see half equivalence point is at p H 5.8 . We know that at half equivalence point p H = p K a = 5.8 . From the p K a value, we can calculate K b value for conjugate base of weak acid and then [ O H - ] of the salt solution using ICE table and K b e q u a t i o n . Once we know the [ O H - ] concentration we can find the p H of the solution. Formula: i) p K a + p K b = 14 ii) p K b = - log K b iii) K b = O H - [ H A ] [ A - ] iv) p O H = - log O H - v) p H + p O H = 14 Given: Initial concentration of sodium salt of weak acid = 0.5 M Figure 15.3 of the textbook. Calculations: Let us assume that the sodium salt of the weak acid is N a A . The hydrolysis reaction of conjugate base A - of weak acid H A is written as: A - a q + H 2 O l ⇌ H A a q + O H - a q Calculating the K b value from the p K a = 5.8 . p K a + p K b = 14 p K b = 14 - p K a p K b = 14 - 5.8 = 8.2 Now, p k b = - log K b K b = 10 - p K b = 10 - 8.2 = 6.31 × 10 - 9 The concentration of the salt N a A is 0.5 M . Thus, the concentration of A - will also be 0.5 M . To determine [ O H - ] we have to draw ICE table for the above reaction R e a c t i o n A - a q + H 2 O l ⇌ H A a q + O H - a q A - M H A M O H - M I n i t i a l 0.5 0 0 C h a n g e - x + x + x E q u i l i b r i u m 0.5 - x x x K b = O H - H A A - 6.30 × 10 - 9 = x 2 ( 0.5 - x ) Because K b is small relative to initial concentration of A - , we can make the simplifying assumption that x is small compared with 0.5 M , and so 0.5 - x ≈ 0.5 6.30 × 10 - 9 = x 2 ( 0.5 ) x 2 = 6.30 × 10 - 9 × 0.5 = 3.15 × 10 - 9 x = O H - = 5.6167 × 10 - 5 M Now calculating p H of the solution from [ O H - ] p O H = - log O H - p O H = - log ⁡ 5.6167 × 10 - 5 M p O H = 4.25 p H + p O H = 14 p H = 14 - 4.25 = 9.75 Therefore, the p H of 0.5 M solution of sodium salt of the weak acid is 9.75 . Conclusion: At half equivalence point, the p H of the solution is equal to the p K a of the weak acid.

Solution Summary: The author calculates the pH of the solution of sodium salt of weak acid using the graph given in Figure 15.3.

Chemistry: An Atoms-Focused Approach

Chemistry: An Atoms-Focused Approach

14th Edition

ISBN: 9780393912340

Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster

Publisher: W. W. Norton & Company

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Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

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Chapter 15, Problem 15.3VP

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Chemistry: An Atoms-Focused Approach

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Acid-Base Titration | Acids, Bases & Alkalis | Chemistry | FuseSchool; Author: FuseSchool - Global Education;https://www.youtube.com/watch?v=yFqx6_Y6c2M;License: Standard YouTube License, CC-BY