# C142_Lec22_Temp_Au-02.ppt

## Chemistry 142 with Callis at University of Washington - Seattle Campus *

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- C142_Lec22_Temp_Au-02.ppt

Michael W.

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pH of Very Weak Acids; (b) pH of Dilute Solutions of Strong Acids; Chemistry 142 B Autumn Quarter 2004 J. B. Callis, Instructor Lecture #22 Acid Solutions in Which Water Contributes to [H+] Previously, we have assumed that the acid is the dominate source of H+ ions. However, in some cases the contribution of H+ ions from water must be taken into account. This situation occurs when Kw/[H+] is comparable to [H+], i.e. near neutral. Examples include low concentrations of weak (or even strong) acids, and acids with pKas near 7. Posing the Weak Acid Problem We have the following four unknowns to solve for: We have the following four equations which govern the concentrations of the above species: The Solution to the Weak Acid Problem We now proceed to solve the four equations for [H+] by successive elimination of variables. The result for [H+] is Finding [H+] requires that we find the three roots of the above cubic equation and then selecting the one root that is consistent with physical reality, i.e. leads to all positive concentrations. Fortunately, in the above case only one root is positive and the other two are negative. This positive root can be found by the ‘Solver’ function of your calculator. You can make a wild guess that is positive and the calculator will converge to the correct answer. Problem 22-1: pH of a Very Weak Acid Find the pH of 3.0 x 10-5 g of vitamin B1, thiamine hydrochloride (HC12H17ON4SCl2, mw = 337.27 g/mol ), a weak acid with Ka = 3.4 x 10-7. The volume is 1.00 L aq. Ans: Problem 22-1: pH of a Very Weak Acid – cont. (a) Start with the expression for the value of [H+] as derived for the case where the autoionization of water cannot be neglected: Now substitute the relevant values for the constants of the problem, Ka, [HA]0 and Kw: Where y = [H+] This expression may be simplified to: Problem 22-1: pH of a Very Weak Acid – cont. (b) To solve: We need to find the physically relevant root. This may be obtained using the ‘Solver’ function of the TI-83 calculator. Solver requires a guess for the answer and boundaries. We know that the desired root must be greater than and less than M. So these can be used as the boundaries. We know that the pH will be near neutral and we have an acid, so we try as the guess for y. We punch in the values and BAM! The answer comes up: [H+] = ; pH = -log( ) = Problem 22-1: pH of a Very Weak Acid – cont. (c) If the autoionization of water is mistakenly neglected, the result is a quadratic equation: Solving gives y = [H+] = M for a pH of . But this pH is on the basic side of 7, which is impossible when an acid is dissolved in water. Strong Acid Solutions in Which Water Contributes to [H+] Usually, a strong acid is the dominant source of H+ ions. However, in some cases the contribution of H+ ions from water must be taken into account. This situation occurs when the concentration of the strong acid is on the order of 10-6 M or less. To solve this problem, we again start with the four equations for the four species. Posing the Strong Acid Problem We have the following four unknowns to solve for: [H+], [OH-], [HA] and [A-] We have the following four equations which govern the concentrations of the above species: The Solution to the Strong Acid Problem We can simplify the four equations, because Ka = infinity, so [HA] = 0, and [A-] = [HA]0, thus we are left with We now proceed to solve the two equations for [H+] by elimination of [OH-]. The result is Finding [H+] requires that we find the two roots of the above quadratic equation and then selecting the one root that is consistent with physical reality, i.e. leads to all positive concentrations. Problem 22-2: pH of a Low Concentration of a Strong Acid Find the pH of 2.0 x 10-8 M solution of HCl. Ans: This is a sufficiently dilute solution that the autoionization of water cannot be neglected. However, since the acid is strong, all of the acid may be assumed dissociated. Thus we need to solve the equation: Where [HA]0 = and Kw = Thus the equation to solve is y = , . We keep the positive root. pH = -log( ) = Answers to Problems in Lecture 22 pH = 6.864 pH = 6.96

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