1) Write the reaction for a diprotic acid, H2A, to reach the first equivalence p

Question

1) Write the reaction for a diprotic acid, H2A, to reach the first equivalence point with NaOH.

2) Write the reaction for a acid HA-, to reach the second equivalence point with NaOH.

3) Write the net equation for the reaction of a diprotic acid, H2A, to reach the second equivalence point with NaOH. (Hint, combine your answers from questions 1 and 2)

4) What species would be present at the first halfway point for the titration of a diprotic acid, H2A, with NaOH. (Hint: look at your answer to question 1 and think about the halfway point).

5) What species would be present at the second halfway point for the titration of a diprotic acid, H2A, with NaOH. (Hint: look at your answer to question 2 and think about the halfway point).

6) 0.400 g of an unknown diprotic acid are reacted to the second equivalence point with 23.2 mL of 0.100 M NaOH. How many moles of diprotic acid were present in the original 0.400 g?

7) What is the molar mass of the unknown diprotic acid in question 6?

*Must answer all parts.. and explain how you got to the answer in order to recieve the points.

Explanation / Answer

1. H2A + NaOH = Na+ HA- + H2O FRIST EQUIVALENT POINT

2. Na+ HA-+ NaOH = ( Na+)2HA-2 + H2O SECOND EQUIVALENT POINT

3 NET EQUATION : ADD EQUATION IN QUESTION 1 AND 2

   H2A + 2 NaOH =  ( Na+)2HA-2 +2 H2O

4 Na+ HA-,  H2O   species would be present at the first halfway point for the titration.

5.  ( Na+)2HA-2, H2O species would be present at the second halfway point for the titration.

6. IN 1000 ml .100 M NaOH .1 MOLE OH- ION IS PRESENT

   IN 23.2 ml .100 M NaOH ( .1/1000)*23.6 = 2.36 *10-3 MOLE OH- ION IS PRESENT

SINCE ONE MOLE OH- ION NUTRALISED ONE MOLE H+ ION

SO H+ ION WAS PRESENT IN SOLUTION IS  2.36 *10-3 MOL

SINCE IT IS DIPROTIC ACID SO,

2 MOLE OF H+ ION WILL OBTAIN FROM 1 MOLE OF DIPROTIC ACID

    2.36 *10-3 MOL OF H+ ION WILL OBTAIN FROM( 1/2)*2.36 *10-3 = 1.18 *10-3 MOLE OF DIPROTIC ACID

1.18 *10-3 MOLE OF DIPROTIC ACID PRESENT

7. MOLE = ( WEIGHT TAKEN / MOLECULAR WEIGHT)

SO ,

MOLECULAR WEIGHT =  ( WEIGHT TAKEN / MOLE)

HERE WEIGHT TAKEN = .400g

MOLECULAR WEIGHT OF UNKNOWN ACID = .400 / 1.18 *10-3 =338.98 gm

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