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Buffers with Specific Ionic Strength

Buffers with Specific Ionic Strength

How many mL of 12.0 M acetic acid and how many grams of sodium acetate (FW = 82 g/mol) are needed to prepare a 500 mL buffer at pH 5.0, and having an ionic strength of 0.2. ka = 1.8*10-5

We need to find the concentration of the salt:m = ½ S CiZi

2

0.2 = ½ (CNa+ * 12 + COAc

- * 12)

CNa+ = COAc

-

0.2 = ½ (2CNa+ )

CNa+ = COAc

- = 0.2 M = CNaOAc

mmol NaOAC = 0.2*500 = 100mg NaOAc = 100*82 = 8200 mg or 8.2 g

2

HOAc D H+ + OAc-

We can now find the concentration of the acid where:

1.8*10-5 = 10-5*0.2/[HOAc][HOAc] = 0.2/1.8 = 0.11 Mmmol HOAc = 0.11*500 = 55.612.0*VHOAc = 55.6

VHOAc = 4.6 mL

3

How many grams of Na2CO3 (FW = 106 g/mol) and how many grams of NaHCO3 (FW = 84 g/mol) are needed to prepare a 1000 mL buffer at pH 10.0, and having an ionic strength of 0.2. ka2 = 4.8*10-11

HCO3- D CO3

2- + H+

4.8*10-11 = 10-10 [CO32-]/[HCO3

-]

[HCO3-] = 2.1[CO3

2-]

m = ½ S CiZi2

4

In NaHCO3, CNa+

= CHCO3-

In Na2CO3, CNa+ = 2CCO3

2-

0.2 = ½ (CNa+ *12 + CHCO3

- *12 + CNa+ *12 + CCO3

2- *22)

0.2 = ½ (CHCO3- *12 + CHCO3

- *12 + 2 CCO32- *12 + CCO3

2- *22)

However, [HCO3-] = 2.1[CO3

2-]

0.2 = ½ (2* 2.1[CO32-] + 2 CCO3

2- *12 + CCO32- *22)

[CO32-] = 0.0392 M

[HCO3-] = 2.1[CO3

2-] = 2.1*0.0392 = 0.0824 M

g Na2CO3 = 0.0392 *1000 * 106= 4.16

g NaHCO3 = 0.0824*1000*84 = 6.92g

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6

How many mL of 12.0 M acetic acid and how many grams of sodium acetate (FW = 82 g/mol) and how many grams of NaNO3 (FW = 85 g/mol) are needed to prepare a 500 mL buffer at pH 5.0, and a salt concentration of 0.1 M and an ionic strength of 0.15 . ka = 1.8*10-5

We need to find the concentration of the salt:m = ½ S CiZi

2

m = {(mNaNO3)+(mNaOAc)}

0.15 = ½ {(CNa+ * 12 + CNO3

- * 12)+ (0.1*12 + 0.1*12)}

CNa+ = CNO3

- = CNaNO3

0.15 = ½ {(2CNaNO3) + 0.2}

CNaNO3 = 0.05 M

g NaNO3 = 0.05*500*85 = 2125mg = 2.125 g

CNaOAc = 0.1 M

mmol NaOAC = 0.1*500 = 50mg NaOAc = 500*82 = 4100 mg or 4.1 g

HOAc D H+ + OAc-

We can now find the concentration of the acid where:

1.8*10-5 = 10-5*0.1/[HOAc][HOAc] = 0.1/1.8 = 0.056 Mmmol HOAc = 0.056*500 = 27.85.0*VHOAc = 27.8

VHOAc = 5.6 mL7