PIS. (Sec. 16.3) A buffer solution is prepared by adding 35.5 mL. of a 0.082 M s
ID: 705172 • Letter: P
Question
PIS. (Sec. 16.3) A buffer solution is prepared by adding 35.5 mL. of a 0.082 M solution of lithi ium benzoate, LiCsH CO2, into 145.5 ml. of 0.17 M benzoic acid, HCaHsCO,. What is the pH of this buffer solution? (Benzoic Acid: Ka-6.5 x10) a) 12.98 b) 8.98 c) 3.26 d) 6.30 e) 11.72 P16. (Sec. 16.3) A buffer solution is prepared by adding KF, to 75 ml of 0.175 M hydrofluoric acid, HF. What is the pH of this buffer solution? (Hydrofluoric Acid: Ka- 7.1 x 10) 35 mL of 0.1 M potassium fluoride, a) 4.42 b) 5.00 c) 3.15 d) 2.57 e) 1.55 P17. (Sec. 16.3) You're asked to prepare a buffer solution with a pH value of 8.75 (pH-8.75) using solutions of hydrocyanic acid, HCN, and sodium cyanide, NaCN. What is the molar ratio, [Base] /[Acid], needed to prepare this buffer solution? (Hydrocyanic Acid: Ka-4.9 x 1010) a) [Base] / [Acid]-363 b) [Base]/[Acid] 0.276 c) [Base]/[Acid]-0.723 d) [Base]/ [Acid] 1.38 e) [Base] /[Acid-0.633 PI8. (Sec. 16. 4) Two Part Question: A buffer solution is prepared by adding 125 mL of 0.35 M the pH of the buffer solution? Part 2: What is the pH of this buffer solution if 75 mL of 0.125 M NaOH is added? (Benzoic Acid: Ka - 6.5 x 10*) a) 5.45; 6.11 b) 2.18;2.43 c) 3.68: 3.34 d) 3.68;4.02 e) 2.18; 1.89 P19. (Sec. 16.4) Two Part Question: A buffer solution is prepared by adding 165 mL of 0.225 M lauric acid, HC12H2302, to 115.5 mL of 0.101 M potassium laurate, KC12H2302. Part 1: What is the pH of the buffer solution? Part 2: What is the pH of this buffer solution if 85.5 M of 0.125 M HCl is added? (Lauric Acid: Ka-501 x 10-6)Explanation / Answer
According to chegg rules we have to answer only first one problem
Buffer solution:
Buffer solution is made by the addition of acid and base solution. The pH of this solution is remain unchanged by adding acid or base. In general the pH of this solution is constant. . A buffer must consist of a weak conjugate acid-base pair, thus it is a weak acid and its conjugate base, or a weak base and its conjugate acid.
pH = pKa + log( Salt /acid)
Ka for benzoic acid --> 6.5*10^-5
pKa = - log Ka = - log 6.5*10^-5
= 4.19
Mole of acid = molarity * volume in L
= 0.17*145.5/1000
= 0.025 moles
Mole of salt = molarity * volume in L
= 0.082 *35.5/1000
= 0002911 moles
Total volume = 145.5+35.5= 181 ml
= 0.181 L
Molarity = number of moles / volumes
Acid: 0.025/ 0.181 = 0.138 M
Salt = 0.002911 / 0.181 = 0.0161 M
Therefore;
pH = 4.19 + log(0.0161/ 0.138)
pH = 4.19 + log
pH = 4.19 + (-0.933)
= 3.26
The correct answer is 3.26.
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