RESULTS Shaded areas should be completed before lab, as you do your pre-lab assi

Question

RESULTS Shaded areas should be completed before lab, as you do your pre-lab assignment Table 1. Literature values for reference (Completed before the lab) Ka Ka Molar mass 36.086 | (74.2 212.2 KH2PO4 .31 ?/2-12 | |2.gy K2HPO4 K3PO4 N/A N/A Table 2. Target buffer solution (Completed before the lab) Target pH value of your buffer solution: (Assigned by your lab TA on D2L) Conjugate acid-base pair that will work best for reaching target pH: NaC Acid: Base: pKg of acid: Chemical equation for this buffer: Buffer ratio needed to reach target pH: (show calculation in space provided on the right) LA-T HA

Explanation / Answer

If it considered as potassium salts specifically (potassium salts have different pKa's than phosphoric acid)

Ka

pKa

Molar mass

KH2PO­4

1.58*10-7

6.82

136.09

K2HPO4

6.31*10-13

12.2

174.2

K3PO4

N/A

N/A

212.27

Target ph value of your buffer solution

(assigned by lab TA)

7.2 (Given)

Conjugate base pair that will work best for reaching target pH – (KH2PO­4 + K2HPO4)

Acid - KH2PO­4

pKa of acid – 6.82

Base - K2HPO4

Chemical reaction for this buffer – H2PO-4 + OH- ? HPO-24 + H3O+

Buffer ration needed to reach target pH – 2.40

[A-]/[HA] =

pH = pKa +log[A-]/[HA]

7.2 = 6.82 +log[A-]/[HA]

0.38 = log[A-]/[HA]

[A-]/[HA] = antilog [0.38]

[A-]/[HA] = 2.398

Approximate number of mole and mass of HA required to make 250 ml of 0.05 M HA solution

nHA =

Mass of potassium salt =

We have to make 0.05 M solution so

HA +A = 0.05

Where as [A-]/[HA] = 2.4

After solving the equations

A = 2.4 HA

HA = 0.0147 mole and A = 0.0353

0.0147 *MW of potassium salt(KH2PO4)

0.0147*136.09 = 2.00g

Approximate number of mole and mass of A required to make 250 ml of 0.05 M HA solution

nA =

Mass of potassium salt =

We have to make 0.05 M solution so

HA +A = 0.05

Where as [A-]/[HA] = 2.4

After solving the equations

A = 2.4 HA

HA = 0.0147 mole and A = 0.0353

0.0353 *MW of potassium salt(K2HPO4)

0.0353*174.2 = 6.149g

If it considered as phosphoric acid

Ka

pKa

Molar mass

KH2PO­4

6.2*10-8

7.21

136.09

K2HPO4

4.8*10-13

12.32

174.2

K3PO4

N/A

N/A

212.27

Target ph value of your buffer solution

(assigned by lab TA)

7.2 (Given)

Conjugate base pair that will work best for reaching target pH – (KH2PO­4 + K2HPO4)

Acid - KH2PO­4

pKa of acid – 6.82

Base - K2HPO4

Chemical reaction for this buffer – H2PO-4 + OH- ? HPO-24 + H3O+

Buffer ration needed to reach target pH – 2.40

[A-]/[HA] =

pH = pKa +log[A-]/[HA]

7.2 = 7.21 +log[A-]/[HA]

0.01 = log[A-]/[HA]

[A-]/[HA] = antilog [0.01]

[A-]/[HA] = 1.023

Approximate number of mole and mass of HA required to make 250 ml of 0.05 M HA solution

nHA =

Mass of potassium salt =

We have to make 0.05 M solution so

HA +A = 0.05

Where as [A-]/[HA] = 1.023

After solving the equations

A = 1.023 HA

HA = 0.0247 mole and A = 0.0253

0.0247 *MW of potassium salt(KH2PO4)

0.0247*136.09 = 3.36g

Approximate number of mole and mass of A required to make 250 ml of 0.05 M HA solution

nA =

Mass of potassium salt =

We have to make 0.05 M solution so

HA +A = 0.05

Where as [A-]/[HA] = 1.023

After solving the equations

A = 1.023 HA

HA = 0.0247 mole and A = 0.0253

0.0253 *MW of potassium salt(K2HPO4)

0.0253*174.2 = 4.407g

Ka

pKa

Molar mass

KH2PO­4

1.58*10-7

6.82

136.09

K2HPO4

6.31*10-13

12.2

174.2

K3PO4

N/A

N/A

212.27

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