A series of solutions are made from 0.10M HA and 0.10M NaA. There are six total

Question

A series of solutions are made from 0.10M HA and 0.10M NaA.

There are six total solutions:

Soultion 1: has a pH of 2.87. It is composed of 50mL 0.10M HA

Soultion 2: has a pH of 3.86. It is composed of 40mL 0.10M HA + 10mL 0.10M NaA

Solution 3: has a pH of 4.13. It is composed of 30mL 0.10M HA + 20mL 0.10M NaA

Solution 4: has a pH of 4.45. It is composed of 25mL 0.10M HA + 25mL 0.10M NaA

Solution 5: has a pH of 4.66. It is composed of 20mL 0.10M HA + 30mL 0.10M NaA

Solution 6: has a pH of 4.97. It is composed of 10mL 0.10M HA + 40mL 0.10M NaA

Explanation / Answer

Soultion 1:)

pH = 2.87

[H+] = 1.35 x 10^-3 M

HA   ----------------> H+ +   A-

0.1                            0        0

0.1 - x                       x         x

[H+] = x = 1.35 x 10^-3 M

Ka = x^2 / 0.1 - x = (1.35 x 10^-3)^2 / 0.1 - 1.35 x 10^-3

Ka = 1.84 x 10^-5

Soultion 2:

millimoles of HA = 40 x 0.1 = 4

millimoles of NaA = 10 x 0.1 = 1

pH = pKa + log [1 / 4]

3.86 = pKa + log 0.25

pKa = 4.462

Ka = 3.45 x 10^-5

Solution 3:

millimoles of HA = 30 x 0.1 = 3

millimoles of NaA = 20 x 0.1 = 2

pH = pKa + log [2 / 3]

4.13 = pKa + log 0.66

pKa = 4.306

Ka = 4.94 x 10^-5

Solution 4:

millimoles of HA = 25 x 0.1 = 2.5

millimoles of NaA = 25 x 0.1 = 2.5

pH = pKa

pKa = 4.45

Ka = 3.55 x 10^-5

Solution 5: has a pH of 4.66. It is composed of 20mL 0.10M HA + 30mL 0.10M NaA

millimoles of HA = 20 x 0.1 = 2

millimoles of NaA = 30 x 0.1 = 3

pH = pKa + log [3/2]

4.66 = pKa + log 1.5

Ka = 3.29 x 10^-5

6)

Ka = 4.29 x 10^-5

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