Question

Titration of hydroglycine chloride, H2GCl. Titrate 20.0 mL of 0.10M H2G+ (pK’s=2.35 and 9.78) with 0.10 M NaOH: Determine pH at VNaOH = 0, 10.0,15.0, 20.0, 25.0, 30.0, and 40.0 mL

Answer 1

1. Glycine hydrochloride with react with NaOH to give glycine in the amine form and you will get the sodium salt of the carboxylic acid end formed as ions.

HO₂CCH₂NH₃Cl + 2NaOH --> Na⁺[O₂CCH₂NH₂]^- + NaCl + 2H₂O

Number of moles of acid in 20 mL of 0.10 M acid = 0.020 0.10 = 0.002 mol acid

No base has been added. Therefore,

i) pKa = 2.35

Ka = 4.47 10^-3 = (0.10 - x)

0.10 - x = 0.10

x² = 4.47 10^-3 0.10

x = 0.021142 = [H⁺]

pH = -log [H⁺] = -log [0.021142] = - (-1.674) = 1.67

ii) pKa = 9.78

Ka = 1.66 10^-10 = (0.10 - x)

0.10 - x = 0.10

x² = 1.66 10^-10 0.10

x = 0.0000040743 = [H⁺]

pH = -log [H⁺] = -log [0.0000040743] = - (-5.39) = 5.39

2. After the addition of 10 mL of NaOH,

Number of moles of NaOH = 0.010 0.10 = 0.001 mol NaOH

Moles of acid in excess = 0.002 - 0.001 = 0.001
Moles glycinate ion formed = 0.001

pH = 2.35 + log ([0.001] [0.001]) = 2.35

pH = 9.78 + log ([0.001] [0.001]) = 9.78

3. After the addition of 15 mL of NaOH,

Number of moles of NaOH = 0.015 0.10 = 0.0015 mol NaOH

Number of moles of base in excess = 0.002 - 0.0015 = 0.0005

Moles glycinate ion formed = 0.0015

Total volume = 0.035 L

[OH^-]= 0.0005 0.035 = 0.0143 M

pOH = - log [OH^-] = - log [0.0143] = - (-1.845) = 1.845

pH = 14 - 1.845 = 12.16

4. After the addition of 20 mL of NaOH,

Number of moles of NaOH = 0.020 0.10 = 0.002 mol NaOH

Moles glycinate ion formed = 0.002

Total volume = 0.040 L

[OH^-]= 0.002 0.040 = 0.05 M

pOH = - log [OH^-] = - log [0.05] = - (-1.301) = 1.301

pH = 14 - 1.3 = 12.7

5. After the addition of 25 mL of NaOH,

Number of moles of NaOH = 0.025 0.10 = 0.0025 mol NaOH

Moles glycinate ion formed = 0.0025

Total volume = 0.045 L

[OH^-]= 0.0025 0.045 = 0.06 M

pOH = - log [OH^-] = - log [0.06] = - (-1.22) = 1.22

pH = 14 - 1.22 = 12.78

6. After the addition of 30 mL of NaOH,

Number of moles of NaOH = 0.030 0.10 = 0.003 mol NaOH

Moles glycinate ion formed = 0.003

Total volume = 0.050 L

[OH^-]= 0.003 0.05 = 0.06 M

pOH = - log [OH^-] = - log [0.06] = - (-1.22) = 1.22

pH = 14 - 1.22 = 12.78

7. After the addition of 40 mL of NaOH,

Number of moles of NaOH = 0.040 0.10 = 0.004 mol NaOH

Moles glycinate ion formed = 0.004

Total volume = 0.060 L

[OH^-]= 0.004 0.06 = 0.07 M

pOH = - log [OH^-] = - log [0.07] = - (-1.55) = 1.55

pH = 14 - 1.55 = 12.45