Lecture+2+homework+solutions

Lecture+2+homework+solutions - Solutions for problems...

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Solutions for problems assigned with Lecture 2 Garrett & Grisham, pg. 45 1. Calculate the pH of the following. a. 5 x 10 -4 M HCl Answer: HC1 is a strong acid and fully dissociates into [H + ] and [C1 - ]. Thus, [H + ] = [C1 - ] = [HC1] total added pH =− log 10 [H + ] =− log 10 [ HCl total ] =− log 10 (5 × 10 4 ) = 3.3 b. 7 x 10 -5 M NaOH Answer : For strong bases like NaOH and KOH, pH = 14 + log 10 [ Base] = 14 + log 10 (7 × 10 5 ) = 9.85 c. 2 μ M HC1 Answer: pH =− log 10 [H + ] =− log 10 [ HCl total ] =− log 10 (2 × 10 6 ) = 5.70 d. 3 x 10 -2 M KOH Answer: pH = 14 + log 10 (3 × 10 2 ) = 12.5 e. 0.04 mM HC1 Answer: pH =− log 10 [H + ] =− log 10 [ HCl total ] =− log 10 (0.04 × 10 3 ) = 4.4 f. 6 x 10 -9 M HC1= 0.06 x 10 -7 M HCl Answer : Beware! Naively one might fall into the trap of simply treating this like another strong acid problem and solving it like so: pH =− log 10 [H + ] =− log 10 [HCl total ] =− log 10 (6 × 10 9 ) = 8.22 However, something is odd. This answer suggests that addition of a small amount of a strong acid to water will give rise to a basic pH! What we have ignored is the fact that water itself will contribute H + into solution so we must consider the ionization of water as well. There are two approaches we can take in solving this problem. As a close approximation we can assume that: [H + ] = 10 7 + [ HCl ] or, [H + ] = 10 7 + 6 × 10 9 = 10 7 + 0.06 × 10 7 = 1.06 × 10 7 pH =− log 10 (1.06 × 10 7 ) = 6.97

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The exact solution uses the ion product of water. [H + ][OH ] = K W = 10 14 (the ion product of water) (1) In solution HC1 fully dissociates into H + + C1 - . For any solution the sum of negative and ing with monovalent ion we can write: positive charges must be equal. Since we are deal [H ] + [OH ] (2) + ] = [Cl Now because HC1 is fully dissociated: [Cl ] = [ HCl total ] = 6 × 10 9 (3) - Substituting this into equation (2), solving equation (2) for [OH ] and substituting in equation (1) we have a quadratic equation in H + : [H + ] 2 6 × 10 9 [H + ] 10 14 = 0 4ac whose general solution is given by [H + ] = b ± b 2 2a = 6
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Lecture+2+homework+solutions - Solutions for problems...

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