Question

Draw the titration curve for the titration of 20 mL of a 0.12 M solution of glutamate (carboxylic acid pKa = 2.19, amino pKa = 9.67, you need to know the pKa for the R group) with 0.10 M sodium hydroxide. Clearly label the axes of your graph (pH vs mL of NaOH added.) Be sure to include pH values and volumes for the following: Start of the titration (no sodium hydroxide added), the pH at the volume that is half of the volume of each endpoint, pH and volume at the end points of the titration. Show all of your calculations

Answer 1

According to the question;

As we know that;

Glutamate is a conjugate base of glutamic acid. Its acidic in nature and it is necessary for body as it act as neurotransmitter.

So,

Basically it is triprotic acid which have -COOH, -NH₃ and R group. This R group also contains protons.

These triprotic acids behave like an zwitterion under physiological conditions.

So,

According to graph we can say that first titration point is at pH = 2.29 and the second titration point is at pH =4 and the third equivalence point is at around9.67.

So,

Now we can determine the pK_a value by the Handerson - Hasselbach equation.

Then,

Firstly we calculate the values for A^- and HA

So,

In this case we are taking first equivalence point is at 24 ml when the pH is 2.29 and second equivalence point is at 48ml when the pH is 4 and third equivalence point is at 72ml when pH is 9,67.

So, finding the volume of amino acid on second equivalence point is basically

Then,

Volume at first equivalnce point at 24 ml = 20ml of amino acid + 24 ml of NaOH = 44 ml

Then,

Volume at second equivalence point at 48 ml = 20 ml of amino acid + 48ml of NaOH = 68ml

So,

Actual volume is= 68 - 44 = 24 ml

Moles of glutamate = 24 /1000 *0.12 = 0.0029

Moles of NaOH = 24 /1000 * 0.10 = 0.0024

So,

Now when glutamate reacts with NaOH = 0.0029 - 0.0024 = 0.0005

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

4 = pKa + log 0.0029 /0.0005

4 = pKa+ log 5.8

4 = pKa + 0.763

pKa= 4 - 0.763

PKa = 3.24

For the R group pKa value is 3.24.