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What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa? Ka = 1.8×10-5 Quantity = mol Submit b What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00? Ka = 1.8×10-5 Quantity = mol c What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00? Ka = 1.8×10-5 Quantity = mol

Respuesta :

Answer:

a) We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

b) We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

c) We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

Explanation:

a) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = pK = -log(1.8*10^-5) = 4.74

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4.74 = 4.74 + log(A-/HA)

0 =  log(A-/HA)

A-/HA = 1

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1

X =0.9

We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4.74 = 4.74 + log(0.9/0.9) = 4.74

b) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 4

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4 = 4.74 + log(A-/HA)

-0.74 =  log(A-/HA)

A-/HA = 0.182

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 0.182

X =0.277

We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4 = 4.74 + log(0.277/1.523)

c) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 5

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

5 = 4.74 + log(A-/HA)

0.26 =  log(A-/HA)

A-/HA = 1.82

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1.82

X =1.16

We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

5 = 4.74 + log(1.16/0.64) = 5

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