Question

The Ka of lactic acid, HC3H5O3, is 1.4 ✕ 10−4.

(a) Suppose buffer #1 is prepared using 40.0 mL 0.1 M HC3H5O3 and 60.0 mL 0.1 M C3H5O3− to give a final volume of 100.0 mL. What is the pH of this buffer? WebAssign will check your answer for the correct number of significant figures.

4.19 Incorrect: Your answer is incorrect.

(b) Suppose a buffer #2 is prepared using 60.0 mL 0.1 M HC3H5O3 and 40.0 mL 0.1 M C3H5O3− to give a final volume of 100.0 mL. What is the pH of this buffer? WebAssign will check your answer for the correct number of significant figures.

5.37 Incorrect: Your answer is incorrect.

(c) Why is buffer #2 more acidic than buffer #1? It has a higher acid concentration. It has a greater volume of base. It was prepared after buffer #1. It has a greater final volume. Correct: Your answer is correct.

(d) Calculate the pH of each buffer if 0.001 moles of OH − are added to each.

Answer 1

classmate Date Page a) Lactic acid, Ctz H Coz Aç Oz .ka, 21.48104 ² pha= -log (1:48104) = 3.85 Now, buffes I was prepared by mixine 40 ml of olM HCg H5 O₂ and 60 im o.IM Cg His Og This [H Cg H50 ] = 40 xo. 1 20.04 - [ C3 HS O ] = 60 x0. 1 0.06 katlog [ 63 4503 "рни Гне на 04) = 3.85+ 109 0.06 = 4:026. 0.04 b) Bufter 2 was prepared asing by mixing come of 0.14 Hog H5Og and 40 ml of 0.14 Cz ltzo - 1 ...[ H Cz Hg Og] = 60 x0.1 = 0.06 [ {z HBO 3 ] = 40 x0.1 = 0104 pka pH= 3.85 + 169 % + 1680 to = 3.67 ( Tus buffer a has lower pH compared to buftes 1. Because, concentration of acid. is higher compared to conjugate base in bufter 2. But in butter Il reverce situation occurs. Thus, buffer f is more oniric

classmate d) Now buffer 1, o ou au Hencltzo hoomi 0.04M, 100ml 4ghts z - o oux 100 =0.004 mohol 1000 HGH5O₂: while, 0.06M, 100 m Catgo 0.068100 = 0.006 mole H50,- After addition of o.ool mol Nao H, mole no of H Cz 1450g = 0.004 -0.001 20.0039 mole no of Cz H50,= 0.006+0.0012 0.006 L * mole no a concentration.o me can use mole number instead of concentration. Ancho HSO T: PH = pkarlog in tez H5Oz 0.0060 = 3.55+ log ovo 039 24.044 Bufter 2 Similarly in buffer 2 - Ho After addi Non of 0.001M Naot, mole no af HCg Hs Og = 0.006-00001=000050 mola no of Cz 450,- so.004 +0°ool-000041 pH = pkatig met C3 Itsz = a 0.0047 3.857107 or 0959 83.692.