Question

Please help, I'm so confused!!!! This is due wednesday night!!! i'm gonna fail :((((
pH of Buffer Solutions Procedure: Acetic Acid-Sodium Acetate Buffer (pKa acetic acid = 4.75) Weigh about 3.5 g of Na2C2H302 3H2O, record the exact mass, and add to a 250 ml beaker. Measure exactly 8.8 mL of 3.0 M acetic acid (use 10 mL grad cylinder) and add to the beaker containing the sodium acetate. • Measure exactly 55.6 mL of distilled water and add to the beaker. Stir until all of the salt is dissolved. Measure the pH of this solution Pour 1/2 of this solution (32 mL) into a 100 mL beaker. Label these beakers 1 & 2. • Pipette 1.0 mL 6.0 M HCl into beaker 1, mix, measure and record the pH. • Pipette 1.0 mL 6.0 M NaOH into beaker2, mix, measure pH. Record Calculate pH values of the original buffer solution, the values after the additions of HCl and NaOH. How do the measured and calculated values compare? Carbonic Acid, Sodium Hydrogen Carbonate Buffer Systems (pKa carbonic acid = 6.37) System 1:1 Place 32.0 mL of carbonic acid in a beaker and add 32.0 mL of 0.1 M sodium hydrogen carbonate to the beaker. Mix. Measure the pH of this solution Pour 1/2 of this solution (32 mL) into a 100 mL beaker. Label these beakers 1 & 2. Pipette 0.1 mL 6.0 M HCl into beaker 1, mix, measure and record the pH. Pipette 0.1 mL 6.0 M NaOH into beaker2, mix, measure pH. Record. System 1:10 Place 6.4 mL of carbonic acid in a beaker and add 64.0 mL of 0.1 M sodium hydrogen carbonate to the beaker. Mix Measure the pH of this solution Pour 1/2 of this solution (35.2 mL) into a 100 mL beaker. Label these beakers 1 & • Pipette 0.1 mL 6.0 M HCl into beaker 1, mix, measure and record the pH. Pipette 0.1 mL 6.0 M NaOH into beaker2, mix, measure pH. Record Water as a "Buffer" Place 32 mL of distilled water into 2 beakers as above. Measure and record the pH in 1 beaker. Add 1.0 mL of HCl to one beaker and 1.0 mL of NaOH to the other. Record and record the pH in each beaker.

Answer 1

Acetic acid-sodium acetate buffer:
Measured pH of original buffer solution = 4.80
1. moles of sodium acetate:
moles, nNaOAc = mass of sodium acetate /molar mass of sodium acetate
nNaOAc = 3.51/136 = 0.0258 moles
Calculated pH of original buffer solution:
Total volume of solution: 55.6 mL water + 8.8 mL 3M acetic acid (HA) = 64.4 mL
2. Calculate M [A-]: Molarity = moles/volume
[A-] = 0.0258/0.0644 = 0.40 M
3. Calculate M [HA]:

using the dilution relation as M1V1 = M2V2
M1 =stock concnetration of HA, V1 = volume of HA in solution, V2 = solution volume
thus [HA] in solution, M2 = M1V1 /V2
[HA] = 8.8*3/64.4 = 0.41 M
4. pKa of acetic acid: 4.74
5. pH = pKa+log ([A-]/[HA])
substituting the values we get:
pH = 4.74 + log(0.40/0.41)
pH = 4.74 - 9.9*10-3
pH = 4.73
Addition of Acid to buffer

Measured pH of buffer + HCl = 4.24
Calculated pH of buffer + HCl solution:
Total volume of solution: 32.2 + 1 mL 6M HCl = 33.2 mL
1. calculate M HCl added: 1*6/33.2 = 0.18 M HCl
2. new[A-] = 0.40 M [A-] - 0.18 M [HCl] = 0.22 M
3. new [HA] = 0.41 M [HA] + 0.18 M [HCl] = 0.59 M
4. pH = pKa+log ([A-]/[HA])
substituting the values we get:
pH = 4.74 + log(0.22/0.59)
pH = 4.74 - 0.43
pH = 4.31
Upon addition of acid, pH of decreases.
Addition of NaOH to buffer

pH of buffer + NaOH = 5.21
Calculated pH of buffer + NaOH solution:
Total volume of solution: 32.2 + 1 mL 6M NaOH = 33.2 mL
1. calculate M NaOH added: 1*6/33.2 = 0.18 M NaOH
2. new[A-] = 0.40 M [A-] + 0.18 M [NaOH] = 0.58 M
3. new [HA] = 0.41 M [HA] - 0.18 M [NaOH] = 0.23 M
4. pH = pKa+log ([A-]/[HA])
substituting the values we get:
pH = 4.74 + log(0.58/0.23)
pH = 4.74 + 0.40
pH = 5.14
Upon addition of base, pH of increases.

Carbonic acid-sodium hydrogen carbonate buffer system

System 1:1

stock solution [HA] = [A-] = 0.1 M

Calculated pH of original buffer solution:
Total volume of solution: 32.0 mL HA + 32.0 mL 0.1M A- = 64.0 mL
1. Calculate M [A-]: using the dilution relation as M1V1 = M2V2
M1 = stock concentration of A-, V1 = volume of A- in solution, V2 = solution volume
thus [A-] in solution = M1V1/V2
[A-] = 0.1*32.0/64.0 = 0.05 M
2. Calculate M [HA]:
for 1:1 system [HA] = [A-] = 0.05 M
3. pKa of carbonic acid: 6.37
5. pH = pKa+log ([A-]/[HA])
substituting the values we get:
pH = 6.37 + log(0.05/0.05)
pH = 6.37 + 0
pH = 6.37

Calculated pH of buffer + HCl solution:
Total volume of solution: 32.0 + 0.1 mL 6M HCl = 32.1 mL
1. calculate M HCl added: 0.1*6/32.1 = 0.0187 M HCl
2. new[A-] = 0.05 M [A-] - 0.0187 M [HCl] = 0.0313 M
3. new [HA] = 0.05 M [HA] + 0.0187 M [HCl] = 0.0687 M
4. pH = pKa+log ([A-]/[HA])
substituting the values we get:
pH = 6.37 + log(0.0313/0.0687)
pH = 6.37 - 0.34
pH = 6.03
Upon addition of acid, pH of decreases.

Calculated pH of buffer + NaOH solution:
Total volume of solution: 32.0 + 0.1 mL 6M NaOH = 32.1 mL
1. calculate M NaOH added: 0.1*6/32.1 = 0.0187 M NaOH
2. new[A-] = 0.05 M [A-] + 0.0187 M [NaOH] = 0.0687 M
3. new [HA] =0.05 M [HA] - 0.0187 M [NaOH] = 0.0313 M
4. pH = pKa+log ([A-]/[HA])
substituting the values we get:
pH = 6.37 + log(0.0687/0.0313)
pH = 6.37 + 0.34
pH = 6.71
Upon addition of base, pH of buffer increases.

System 1:10 [A-] = 10[HA]

stock solution [HA] = [A-] = 0.1 M

Calculated pH of original buffer solution:
Total volume of solution: 6.4 mL HA + 64 mL 0.1M A- = 70.4 mL
1. Calculate M [A-]: using the dilution relation as M1V1 = M2V2
M1 = stock concentration of A-, V1 = volume of A- in solution, V2 = solution volume
thus [A-] in solution = M1V1/V2
[A-] = 0.1*64/70.4 = 0.091 M
2. Calculate M [HA]:
for 1:10 system [HA] = 0.1* [A-] = 0.0091 M
3. pKa of carbonic acid: 6.37
5. pH = pKa+log ([A-]/[HA])
substituting the values we get:
pH = 6.37 + log(0.091/0.0091)
pH = 6.37 + 1
pH = 7.37

Calculated pH of buffer + HCl solution:
Total volume of solution: 35.2 + 0.1 mL 6M HCl = 35.3 mL
1. calculate M HCl added: 0.1*6/35.3 = 0.017 M HCl
2. new[A-] = 0.091 M [A-] - 0.017 M [HCl] = 0.074 M
3. new [HA] = 0.0091 M [HA] + 0.017 M [HCl] = 0.108 M
4. pH = pKa+log ([A-]/[HA])
substituting the values we get:
pH = 6.37 + log(0.074/0.108)
pH = 6.37 - 0.16
pH = 6.21
Upon addition of acid, pH of decreases.

Calculated pH of buffer + NaOH solution:
Total volume of solution: 35.2 + 0.1 mL 6M NaOH = 35.3 mL
1. calculate M NaOH added: 0.1*6/35.3 = 0.017 M NaOH
2. new[A-] = 0.091 M [A-] + 0.017 M [NaOH] = 0.108 M
3. new [HA] =0.0091 M [HA] - 0.017 M [NaOH] = 0.074 M
4. pH = pKa+log ([A-]/[HA])
substituting the values we get:
pH = 6.37 + log(0.108/0.074)
pH = 6.37 + 0.16
pH = 6.53
Upon addition of base, pH of increases.

Water as buffer:

[H2O] = 55.55 M

In water, [OH-] = [H+]10-7 M

pH = -log [H+]

pH = 7

Addition of 1 mL of 6 M HCl

Volume of solution = 32 mL water + 1 mL of 6 M HCl = 33 mL

[H+] due to 1 mL of 6 M HCl = 1*6/33 = 0.182 M

total [H+] = 10-7 M + 0.182 M = 0.182 M

pH = -log (0.182)

pH = 0.74

pH of water decreases upon addition of NaOH.

Addition of 1 mL of 6 M NaOH

Volume of solution = 32 mL water + 1 mL of 6 M NaOH = 33 mL

[OH-] due to 1 mL of 6 M naOH = 1*6/33 = 0.182 M

total [OH-] = 10-7 M + 0.182 M = 0.182 M

pOH = -log (0.182)

pOH = 0.74

using the formula to relate pH and pOH

pH = 14 - pOH

pH = 14 -0.74

pH = 13.26

pH of water increases upon addition of acid.