A buffer is prepared by adding 65.5 mL of 0.175 M sodium acetate, NaC_2H_3O_2, t
ID: 1063602 • Letter: A
Question
A buffer is prepared by adding 65.5 mL of 0.175 M sodium acetate, NaC_2H_3O_2, to 175 mL of 0.025 M acetic acid, HC_2H_3O_2. What is the pH of this buffer solution? (Acetic Acid: K_a. = 1.8 times 10^-5) 2.32 5.42 4.21 9.18 11.80 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: K_a = 4.9 times 10^-10) [Base]/[Acid] = 3.63 [Base]/[Acid] = 0.276 [Base]/[Acid] = 0.723 [Base]/[Acid] = 1.38 [Base]/[Acid] = 0.633 You're asked to prepare a buffer solution with a pH value of 4.00 (pH = 4.00) using solutions of formic acid, HCOOH, and potassium formate, KCHOO. What is the molar ratio, [Base]/[Acid], needed to prepare this buffer solution? (Formic Acid: K_a = 1.7 times 10^-1) [Base]/[Acid] = 8.41 [Base]/[Acid] = 1.70 [Base]/[Acid] = 0.428 [Base]/[Acid] = 1.95 [Base]/[Acid] = 2.17 Two Part Question: A buffer solution is prepared by add 125 mL of 0.35 M benzoic acid, HC_7H_5O_2, to 125 mL of 0.111 M sodium benzoate. NaC_7H_5O_2. Part 1: What is the pH of this buffer solution? Part 2: What is the pH of this buffer solution if 75 ml. of 0.125 M NaOH is added? (Benzoic Acid: K_a = 6.5 times 10^-5) 5.45; 6.11 2.18:2.43 3.68; 3.34 3.68; 4.02 2.18; 1.89 A buffer solution is prepared by add 165 mL of 0.225 M lauric acid. HC_12H_23O_2, to 115.5 mL of 0.101 M potassium laurate, KC_12H_23O_2. Part 1 What is the pH of this buffer solution? Part 2: What is the pH of this buffer solution if 85.5 mL of 0.125 M HCl is added? (Lauric Acid: K_a = 5.01 times 10^-6)Explanation / Answer
10. pka of aceticacid = -log(1.8*10^-5) = 4.74
pH Of acidic buffer = pka + log(salt/acid)
no of mole ofaceticacid = 175*0.225 = 39.375 mol
no of mole of acetate (salt) = 65.5*0.175 = 11.46 mol
pH = 4.74 + log(39.375/11.46)
= 5.27
answer: b
11.
pka of HCN = -log(4.9*10^-10) = 9.31
pH Of acidic buffer = pka + log(BASE/acid)
8.75 = 9.31 + logx
x = BASE/acid = 0.275
ANSWER: B
12.
pka of HCN = -log(1.7*10^-4) = 3.77
pH Of acidic buffer = pka + log(BASE/acid)
4 = 3.77 + logx
X = 1.7 = BASE/acid
ANSWER: B
13.
pka of BENZOICacid = -log(6.5*10^-5) = 4.19
pH Of acidic buffer = pka + log(salt/acid)
no of mole ofBENZOICacid = 125*0.35 = 43.75 mol
no of mole of BENZOATE (salt) = 125*0.111 = 13.875 mol
pH = 4.19 + log(13.875/43.75)
= 3.7
no of mole of NaOH added = 75*0.125 = 9.375 mole
pH Of acidic buffer = pka + log(salt+NaOH/acid- NnaOH)
ph = 4.19 + log((13.875+9.375)/(43.75-9.375))
= 4.02
ANSWER: d
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.