The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F. b. Calculate the pH at the halfway points to equivalence, B and D. [Key

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1.

Major Species at B:

1.

2.

Major Species at C:

1.

Major Species at D:

1.

2.

Major Species at E:

1.

Major Species at F:

1.

b) pH calculation:

At Halfway point B:

pH = pK + log[]/[H]

pH = pK = 6.35

Similarly, at halfway point D.  

At point D,

pH = pK + log [H]/[H2]

pH = pK = 10.33

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.