Question

How do you get the 1.18 when doing the quadratic equation?

Therefore, we have pH = 4.74 + log(1.00 times 10^-4)/(1.00 times 10^-3) pH = 4.74 - 1.00 = 3.74 Therefore Now we will redo the calculation, without making any assumptions at all. We will solve for x, which represents the H^+ concentration at equilibrium, in order to calculate pH. [CH_3COO^-][H^+]/[CH_3COOH] = K_a (1.00 times 10^-4 + X)(X)/(1.00 times 10^-3 - X) = 1.8 times 10^-5 1.00 times 10^-4 X + X^3 = 1.8 times 10^-5 (1.00 times 10^-3 - X) X^2 + 1.00 times 10^-4 X = 1.8 times 10^-8 = 1.8 times 10^-5 X X^2 + 1.18 times 10^-4 X - 1.8 times 10^-4 = 0 X = -1.18 times 10^-4 plusminus Squareroot (1.18 times 10^-4)^2 - 4(1) (-1.8 times 10^-4)/2(1) = -1.18 times 10^-4 plusminus Squareroot 8.5924 times 10^-4/2 = 8.76 times 10^-5 = [h^+] Ph = 4.06 In Sample Exercise 17.3 the calculate pH is the same whether we solve exactly using the quadratic equation or make the simplifying assumption that the equilibrium concentrations of acid and base are equal to their initial concentrations. The simplifying assumption works because the concentrations of the acid base conjugate pair are both a thousand times larger then k_a in this Sample Exercise, the acid-base conjugate pair concentrations are only 10-100 as large as K_p. Therefore, we cannot assume that x is small compared to the initial concentrations (that is that the small concentrations are essentially equal to the equilibrium concentrations). The best answer to this Sample Exercise is pH = 4.06. obtained without assuming x is small. Thus we see that the assumptions behind the Henderson-Hasselbalch equation become less accurate the acid/base becomes stronger and/or its concentration becomes smaller. A buffer is made with sodium acetate (CH_3COONa) and acetic acid (CH_3COOH); the for acetic acid is 1.80 times 10^-5. The pH of the buffer is 3.98. What is the ratio of the equilibrium concert ration of sodium acetate to that of acetic acid? -0760, 0.174, 0.840, 5.75, Not enough information it given to answer this question. Calculate the final equilibrium pH of a buffer that initially contains 6.50 times 10^-4 M HOCl and 7.50 times 10^-4 M NaOCl. The K_a of HOCl is 3.0 times 10^-5 In Sample Exercise 17.3 we calculated the pH of a buffered solution. Often we will need to work in the opposite direction by calculating the amounts of the acid and to conjugate base needed to achieve a specific pH. This calculations is illustrated in Sample Exercise 17.5.

Answer 1

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