0.1 M solutions of: HC2H302 HCl NaHSO4 Na2SO4 NaC2H302 NaH2PO4 Na2HPO4 NH4Cl NH3

Question

0.1 M solutions of: HC2H302 HCl NaHSO4 Na2SO4 NaC2H302 NaH2PO4 Na2HPO4 NH4Cl NH3 NaHCO3. Na2CO3 PROCEDURE Part I. Buffer Capacity Work in pairs) 1. Acid added. Place the following amounts of reagents into three clean test tubes. (Note: use 1.0 M HC2H302, not 0.10 M or 6.0 M!) #1 #2 grem #3 Selleu Test Tube 9.0mL 5.0mL 1.0 M HC2H302 1.0mL 1.0mL Yenow acid 5.0mL 1.0 M NaC2H302 9.0mL blue basu. green blue *Add two drops of bromcresol green indicator to each solution. Mix them well. Record the colors. (The acid form of bromcresol green is yellow and the base form is blue.) To each solution add 6.0 M HCl drop by drop, swirling to mix after each drop. Count the number of drops needed to turn each solution yellow. 2. Base added. Repeat the solution preparation as in Step 1, including the added indicator. To each solution add 6.0 M NaOH drop by drop, swirling to mix after each drop. Count the number of drops needed to turn each solution blue. (Solution #3 is usually blue, or nearly so

Explanation / Answer

Answer:

[AcOH] = [AcONa] = 1.0 M ------- (asked to use)

In each trial only volume of each componant used to prepare buffere is changed.

As molarity of both acid nad base componant is same, concentraton ratio of each componant will be proportional to their volume ratio,

Hence,

[AcO-] / [AcOH] = Volume of AcONa / Volume of AcOH

For TT 1 : [AcO-] / [AcOH] = 9.00 / 1.00 = 9 : 1.

For TT 2 : [AcO-] / [AcOH] = 5.00 / 5.00 = 1 : 1

For TT 3: [AcO-] / [AcOH] = 1.00 / 9.00 = 1 : 9

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Anwer : Calculation of pH of solution in each trial.

pH of Buffer is given by Henderson-Hasselbalch equation,

pH = pKa + log ([Base] / [Acid])

For AcOH-AcONa buffer,

pH = pKa + log([AcO-] / [AcOH])

pKa of Acetic acid = 4.75

WIth this above equation takes form,

pH = 4.75 + log([AcO-] / [AcOH]) ------------(1)

Using this equation and above [AcO-] / [AcOH] determined ratio foreach trial,

For TT 1 :

pH = 4.75 + log (9/1)

pH = 4.75 + 0.95

pH = 5.70.

For TT 2:

pH = 4.75 + log(1/1)

pH = 4.75 + 0

pH = 4.75

For TT 3:

pH = 4.75 + log (1/9)

pH = 4.75 - 0.95

pH = 3.80

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3) Best Buffer is the one which do not allow change in pH on addition of strong Acid or Base.

i) The Buffer best against Acid is the one which consumes more # of Acid HCl drops to show change in pH (Indicator Brocresol). Till 69 drops Buffer in TT1 shows no colour change i.e. pH is maintained Basic even on addition of Strong acid HCl and hence Buffer in TT1 act best buffere against acid.

ii) On the same ground we can say that Buffer in TT 3 act as best Buffer against Base added.

iii) But The overall best buffer is the one which responds equally to both addition of Acid and Base.

And the Buffer in TT2 is though for small range but equally resist pH change on addition of acid or Base.

Buffer in TT2 can resist pH change till 32 drops of Acid and 21 drops of Base.

Buffer in TT2 is Best Overall Buffer.

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4) Results are explained in answer of Q3 only.

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