Phosphoric acid is a diprotic acid, and a 0.20m 50.0 ml sample is titrated with
ID: 868368 • Letter: P
Question
Phosphoric acid is a diprotic acid, and a 0.20m 50.0 ml sample is titrated with 12.5 ml of 0.100 m potassium hydroxide to reach the first equivalence point, where the first acid dissociation constant (ka1) = 7.25x10-3 m at 25 degrees c. an additional 12.5 ml, 0.100 m of koh is added to reach the second equivalence point, where the second acid dissociation constant is ka2 = 6.31x10-8 at 25 degrees celsius.
a) Write the balanced chemical equation including all phases of the first dissociate of phosphoric acid.
b) Calculate the ph at the equivalence point.
c) Write the balanced chemical equation including all phases of the second dissociation of phosphoric acid.
d) Calculate the pH at the second equivalence point.
Explanation / Answer
a) balanced equation
H3PO3 (aq) + KOH (aq)-------------------------> KH2PO3 (aq) + H2O (l)
0.2x 50 12.5 x 0.1 0 0
10 1.25 0 0 ---------------------> initial millimoles
8.75 0 1.25 2.25 -------------------> after reaction millilmoles
in the solution Acid and salt only remains . this can form buffer
pKa = -log Ka = -log(7.25x10-3) = 2.14
For acidic buffer
Henderson-Hasselbalch equation
pH = pKa + log[salt/acid]
pH = 2.14 + log[1.25/8.75]
pH= 1.29
b) pH at first equivalence point = 1.29
c) balanced equation for second dissociation
KH2PO3 (aq) + KOH (aq) --------------------------> K2HPO3 (aq) + H2O (l)
1.25 1.25 0 0
0 0 1.25 1.25
here only salt remained
slat millimoles = 1.25
salt concentration = 1.25/total volume = 1.25/(50+12.5+12.5) = 1.25/100= 0.0125
it is salt of strong base and weak acid so pH > 7
pH = 7 + 1/2 [pKa2 + logC]
= 7+ 1/2[ 7.2 + log(0.0125)]
= 7 + 1/2 [7.2 -1.9]
= 9.64
pH at the second equivalence point = 9.64
note : check the acid is phosphoric acid or phosphorous acid.
phosphoric acid tribasic acid (H3PO4) , phosphorous acid dibasic acid (H3PO3).thats why i took phosphorous acid.