deionized with a pH meter and and dry water (out of one beaker onlyl. Be sure to

Question

deionized with a pH meter and and dry water (out of one beaker onlyl. Be sure to rinse the tip of the pH meter with d water it between uses. With a clean, disposable pipette, add 5 drops ofo50 M HCI to one of the beakers containing your buffered solution and 5 ofo50MNaoH to the other beaker containing your buffered solution. Stir the solutions and record the resulting pH Repeat this procedure with the two samples of deionized water Exchange data with group members who made the other three buffer solutions and record directly into your lab notebook. Part B: Designing a Buffer Prepare o.100Lof your assigned buffer in a 100 mL volumetric flask. Using a graduated cylinder measure the necessary volumes of acid, base and water which you have calculated before coming to lab. Chemicals used: 0.50 M HCH,0, 0.50 M NacH,0, 0.50 M HCooH, 0.50 MNacooH Buffer 1: Prepare 0.100 Lof a pH 3.74 buffer with a total buffer strength of 0.100 M. Buffer 2: Prepare 0.100 Lof a pH 404 buffer with a total buffer strength of0.100 M. Buffer 3: Prepare 0.100 Lof a pH 4.44 buffer with a total buffer strength of0.100 M. Buffer 4: Prepare 0.100 L of a pH 5.04 buffer with a total buffer strength of 100 M. 262

Explanation / Answer

Preparation of buffer

Part B)

Buffer 1 : 0.1 L of buffer pH = 3.74, buffer strength = 0.1 M

Using hendersen-Hasselbalck equation,

pH = pKa + log(base/acid)

3.74 = 3.75 + log([HCOONa]/[HCOOH])

[HCOONa] = 0.98[HCOOH]

[HCOOH] + [HCOONa] = 0.1 M x 0.1 L = 0.01 mol

[HCOOH] + 1[HCOOH] = 0.01

[HCOOH] = 0.01/1 = 0.01 mol

volume of 0.5 M HCOOH = 0.01 mol/0.5 M = 0.02 L

So we would take 0.02 L of HCOOH and dilute to 0.1 L to get a buffer with pH 3.74

Buffer 4 : 0.1 L of buffer pH = 5.04, buffer strength = 0.1 M

Using hendersen-Hasselbalck equation,

pH = pKa + log(base/acid)

5.04 = 4.74 + log([NaC2H3O2]/[HC2H3O2])

[NaC2H3O2] = 2[HC2H3O2]

[HC2H3O2] + [NaC2H3O2] = 0.1 M x 0.1 L = 0.01 mol

[HC2H3O2] + 2[HC2H3O2] = 0.01

[HC2H3O2] = 0.01/2 = 0.005 mol

volume of 0.5 M HCOOH = 0.005 mol/0.5 M = 0.01 L

[NaC2H3O2] = 0.01 - 0.005 = 0.005 mol

volume of 0.5 M NaC2H3O2 = 0.005 mol/0.5 M = 0.01 L

so mixing 0.01 L each of 0.5 M acid and conjugate base of acetic acid and diluting with 0.08 L H2O gives a buffer of pH 5.04

Similarly other buffer solutions can be prepared.

Questions,

1. Relation between concentration of buffer and buffer strength

Higher the buffer concentration greater would be the strength of the buffer.

2. Ka expression for

acetic acid

Ka = [NaC2H3O2][H3O+]/[HC2H3O2]

formic acid

Ka = [NaCOOH][H3O+]/[HCOOH]

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