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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 18 - Problem 18.79
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 18 - Problem 18.79

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# Calculate the emf of the following concentration cell at

ISBN: 9780321809247 1

## Solution for problem 18.79 Chapter 18

Chemistry: A Molecular Approach | 3rd Edition

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Problem 18.79

Calculate the emf of the following concentration cell at 25C: Cu(s) 0 Cu21(0.080 M) 0 0 Cu21(1.2 M) 0 Cu(s)

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Chemistry of Solutions Lecture 11 Tuesday, October 4, 2016 Exercise Continued: calculate the pH of a 1.00x10 M H SO solution, for2the 4econd dissociation of H SO2. 4 The major species are H + , HSO - , and H O . (aq) 4 (aq) 2 (l) -2 -3 K a 1.2 x 10 , don't approximate because K must be at aeast 10 smaller than the initial concentration of H SO 2 4 1.2x10 = [HSO ] [H ]4/ [H SO ] 2 4 -2 -2 -2 1.2x10 = (1.0x10 + x) (x) / (1.0x10 – x) (1.2x10 )(1.0x10 – x) = (1.0x10 + x ) -2 2 Using the Quadratic Equation: a = 1, b = 2.2x10 , c = -1.2x10 . -4 -2 -4 Δ = (2.2x10 )2 – 4(1)(-1.2x10 ) -4 Δ = 9.64x10 . (9.64x10 ) -4 1/2= 3.1x10 .-2 -2 -2 -3 + x1= (-2.2x10 + 3.1x10 ) / (2) = 4.5x10 = [H ]. pH = -log(4.5x10 ) = 1.84. - The second dissociation didn't affect the pH of the solution, but the first did. - pH is a function of the 1 dissociation; - if the concentration of an acid is very low, you must consider the first and second K values; a + - The second K doesa't affect the concentration of H enough to change the pH. - Strong acids have low dissociation constants (low K ). The value of a decreases as the concentration of the acid increases, and x increa

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