Buffers are solutions designed to resist changes in pH from the addition of smal

Question

Buffers are solutions designed to resist changes in pH from the addition of small amounts of acids or bases. Buffers are comprised of a solution of a weak acid with its conjugate base. When an outside base is added to the buffer, the weak acid in the buffer neutralizes the hydroxide ion of the base, thus retarding its ability to raise the solution's pH. When an outside acid is added to the buffer, the conjugate base of the buffer neutralizes the hydronium ion of the acid thus retarding the acid's ability to lower the solution's pH. The pH of human blood is maintained near 7.40 by the regulating power of an H_2 CO_3/HCO_3^- buffering system. In today's exercise, you will make buffer solutions of various pH and test their ability to resist the effects of adding strong acid and strong base. Develop an organized data sheet to record the data you collect. The neatness and order of this data sheet will weigh heavily in your grade for this exercise. Prepare 200-mL of 0.010M NaOH solution using the 1.0M NaOH solution provided. Show your calculations for this step in the space provided on the last sheet. Prepare 200-mL of 0.010M HCl solution by diluting the 1.0M HCl solution provided. Show your calculations for this step in the space provided on the last sheet. Prepare 100-mL of 0.010M acetic acid 1.0M HAc solution provided. Show your calculations for this step in the space provided on the last sheet. Prepare 100-mL of 0.010M NH_4 OH solution by diluting the more concentrated ammonia solution provided. Show your calculations for this step in the space provided on the last sheet. Prepare 250-mL of 4.55 pH buffer that is 0.100M in acetic acid using sodium acetate as the conjugate base. (Show your calculations on the last sheet.)

Explanation / Answer

1) Use dilution law: M1 x V1 = M2 x V2

M1 and V1 are molarity and volume of concentrated solution.

M2 and V2 are molarity and volume of diluted solution.

M1 = 1.0 M

V1 = ?

M2 = 0.010 M

V2 = 200 mL

V1 = (M2 x V2)/ M1 = (0.010 M x 200 mL)/ 1.0 M = 2.00 mL

Do the same method for question numbers 2, 3, and 4.

5)

pH of the buffer is

pH = pka + log {[conjugate base]/[acid]}

pH is 4.55 and pKa for acetic acid is 4.76.

4.55 = 4.76 + log {[conjugate base]/(0.100)}

log {[conjugate base]/(0.100)} = -0.21

{[conjugate base]/(0.100)} = 0.62

[conjugate base] = 0.062 M

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