14 Useful constant: K of water at 25 \"C- 1.00 10 L. Data Handling Analysis of t

Question

14 Useful constant: K of water at 25 "C- 1.00 10 L. Data Handling Analysis of the mass percentage of Cl in a given sample gave the following results: 0.582,0.619,0.621 Answer the following questions for the results obtained. Calculate the mean, median, standard deviation, variance, standard error of the mean, range, coefficient of variation, and the 90% confidence limit of the mean. (41 pts) Calculate the absolute and the percentage relative errors of the mean of the data. Assume the true value of the data is 0.62. (12 pts) a. b. II.Acid-Base Equilibria 1. a. Answer either question i or ii.(15 pts) At 25 , calculate the [H'], [OH], pH, and poll of a 0.0800 M solution of a weak monoprotic acid HA with K,-3.0x 10 i. ii. At 25 °C, calculate the [H'], [OH], pH, and pOH of a 0.200 M solution ofa weak monoprotic base B with Kb = 4.35 x 10". b. Answer either question ior i.(20 pts i At 25 , calculate the [H+], [OH'], pH, and pOH of a 0.0860 M solution NaA (the sodium salt of a weak monoprtoic acid HA). Ka of HA is 1.75 x 10 ii. At 25 , calculate the [H'], [OH], pH, and pOH of a 0.0500 M solution BHCI (the salt of a weak monoprotic base B with HCI). Kb of B is 3.2 x 105 2. Answer either question a or b. (30 pts) Calculate the pH of a buffer solution that is 0.100 M in acetic acid, CH CO2H and 0.50 M in its sodium salt, NaCH CO2. Ka of acetic acid is 1.75 x 10 a- ii. Calculate the pH upon adding 0.50 mL of 0.05 M HCI to 5.0 mL of solution in i ili. Calculate the pH upon adding 0. 50 mL of 0.050 M NaOH to 5.0 mL solution in i

Explanation / Answer

a)

pH of weak acid = 1/2(pKa - logC)

pH = 1/2(-log(3*10^-8)-log(0.08))

pH = 4.3099

-log[H+] = 4.3099

[H+] = 10^-4.3099 = 4.8989*10^-5

pOH = 14 - pH = 14 - 4.3099 = 9.6901

pOH = 9.6901

[OH-] = 10^-9.6901

[OH-] = 2.0413 * 10^-10

pOH = 1/2(pKb -logC)

pOH = 1/2(-log(4.35 * 10^-4) - log(0.2))

pOH = 2.0302

[OH-] = 10^-2.0302 = 9.328 *10^-3

pH = 14 - pOH = 14 - 2.0302 = 11.9698

[H+] = 10^-11.9698 = 1.072 * 10^-12

b)

pH of weak acid ands tronf base salt = 7 + 1/2(pKa + logC)

pH = 7 + 1/2(-log(1.75*10^-5) + log(0.086))

pH = 8.8457

-log[H+] = 8.8457

[H+] = 10^-8.5487 = 2.826*10^-9

pOH = 14 - pH = 14 - 8.8457 = 5.1543

pOH = 5.1543

[OH-] = 10^-5.1543

[OH-] = 7.009 * 10^-6

the given compound is formed by weakbase and strong acid,

pH = 7 - 1/2(pKb +logC)

pH = 7 - 1/2(-log(3.2 * 10^-5) + log(0.05))

pH = 5.403

[H+] = 10^-5.403 = 3.954 *10^-6

pOH = 14 - 5.403 = 8.597

[OH-] = 10^-8.597 = 2.529 * 10^-9

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