Question

A solution of 0.146 M cysteine is titrated with 0.0487 M HNO3. The pKa values for...

A solution of 0.146 M cysteine is titrated with 0.0487 M HNO3. The pKa values for cysteine are 1.70, 8.36, and 10.74, corresponding to the carboxylic acid group, thiol group, and amino group, respectively. Calculate the pH at the equivalence point.

Homework Answers

Answer #1

Cysteine titration would have three equivalence points

pKb1 = 3.26

pKb2 = 5.64

pKb3 = 12.3

Ist equivalence point,

pH = 1/2(pKb1 + pKb2)

     = 1/2(3.26 + 5.64)

     = 4.45

IInd equivalence point,

pH = 1/2(pKb2 + pKb3)

     = 1/2(5.64 + 12.3)

     = 8.97

IIIrd equivalence point,

Since the volume of cysteine solution is not mentioned, we cannot calculate the final volume and thus the IIIrd equivalence point.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A solution of 0.243 M cysteine is titrated with 0.0608 M HNO3. The pKa values for...
A solution of 0.243 M cysteine is titrated with 0.0608 M HNO3. The pKa values for cysteine are 1.70, 8.36, and 10.74, corresponding to the carboxylic acid group, thiol group, and amino group, respectively. Calculate the pH at the equivalence point.
A solution of 0.188 M cysteine is titrated with 0.0752 M HNO3. The pKa values for...
A solution of 0.188 M cysteine is titrated with 0.0752 M HNO3. The pKa values for cysteine are 1.70, 8.36, and 10.74, corresponding to the carboxylic acid group, thiol group, and amino group, respectively. Calculate the pH at the equivalence point.
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH....
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (pKa1) and 7.172 (pKa2). A) Calculate pH at the first equivalence point. B) Calculate pH at the second equivalence point.
1) If 25.2 mL of 0.109 M acid with a pKa of 5.55 is titrated with...
1) If 25.2 mL of 0.109 M acid with a pKa of 5.55 is titrated with 0.102 M NaOH solution, what is the pH of the titration mixture after 13.7 mL of base solution is added? 2) If 28.8 mL of 0.108 M acid with a pKa 4.15 is titrated with 0.108 M NaOH solution, what is the pH of the acid solution before any base solution is added? 3)If 27.9 mL of 0.107 M acid with a pKa of...
A volume of 70.0 mL of a 0.860 M HNO3 solution is titrated with 0.450 M...
A volume of 70.0 mL of a 0.860 M HNO3 solution is titrated with 0.450 M KOH. Calculate the volume of KOH required to reach the equivalence point.
The pKa of hypochlorous acid is 7.530. A 51.0 mL solution of 0.117 M sodium hypochlorite...
The pKa of hypochlorous acid is 7.530. A 51.0 mL solution of 0.117 M sodium hypochlorite (NaOCl) is titrated with 0.336 M HCl. Calculate the pH of the solution a) after the addition of 6.45 mL of 0.336 M HCl. b)at the equivalence point with 0.336 M HCl.
The pKa of hypochlorous acid is 7.530. A 54.0 mL solution of 0.147 M sodium hypochlorite...
The pKa of hypochlorous acid is 7.530. A 54.0 mL solution of 0.147 M sodium hypochlorite (NaOCl) is titrated with 0.253 M HCl. Calculate the pH of the solution a) after the addition of 12.5 mL of 0.253 M HCl. pH= ? b) after the addition of 33.5 mL of 0.253 M HCl. pH=? c) at the equivalence point with 0.253 M HCl. pH=?
A volume of 90.0 mL of a 0.820 M HNO3 solution is titrated with 0.220 M...
A volume of 90.0 mL of a 0.820 M HNO3 solution is titrated with 0.220 M KOH. Calculate the volume of KOH required to reach the equivalence point. Express your answer to three significant figures and include the appropriate units.
A 40.0 mL sample of 0.150 M Ba(OH)2 is titrated with 0.400 M HNO3. Calculate the...
A 40.0 mL sample of 0.150 M Ba(OH)2 is titrated with 0.400 M HNO3. Calculate the pH after the addition of the following volumes of acid. a) 0.0 mL b) 15.0 mL c) At the equivalence point d) 40.0 mL
Consider a 0.373 L solution of the amino glutamic acid (0.685 M), which has a carboxylic...
Consider a 0.373 L solution of the amino glutamic acid (0.685 M), which has a carboxylic acid group (pKa = 2.10), an amine group (pKa = 9.47), and a carboxylic acid side chain (pKa = 4.10). How many liters of 2.87 M NaOH would you need to add to reach the isoelectric point of the amino acid?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT