Question

Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture.

???1	???2
1.30	6.70

Calculate the pH for each of the points in the titration of 50.0 mL of 2.4 M H3PO3(aq) with 2.4 M KOH(aq).

OSA —I но—

before addition of any KOH:

after addition of 25.0 mL KOH:

after addition of 50.0 mL KOH:

after addition of 75.0 mL KOH:

after addition of 100.0 mL KOH:

Answer 1

sine molarity of kon added is same as initial molarity of acid. so, at 25.0mL KOH added milinder as one (mu H₂POg converted to the po = 28 x 24 = 60 mmole and then" mmoles of the og still left = 2.4x50-60% = 6ommoles So, at 25 mL kol added : [H₂PO₂ ] = [H₂ PO;] . That is halfway to first equivalence has been reached At som kom, ist equivalence point would be reached when almost all H₂PO, would have converted to H₂POg. At 75mL KOH, midway between test and and equivalence point - 450+25)m occurs and here [H₂POg] = [PO]. At 100 mL KOH second equivalence point occurs and almost all typOg would be in PO form. point Now, for any acid HA at half equivalence when CHATS [A Ke= EATCH O tas [A-J=[HA] se na TĦA) =) [Hot] = ka » - log CH₂ot]= -logka ) pH = pka So, at 25mb, ples pkg = (1-29] ; at 75mL, pH = pha= [6170) At v=ome H₂po₂ (92) + +6014) + H₂POg (as) + Hot (az), k lopta, ni - 0 10 30 Let at equilibrium: Q.4-01) (20)(21) = x² = 21220.05a-0.12=0 12.4- 2.Cobtained from toos . [H₂pQ] using quadra > n = -0.05 +0.0527 YX0:12 = 0.323 Tchot] root formula) Thus, pH= -log*30*) = -log(0.323)=10.49] Initially at v= At 50m, Ist equivalence point: all typOg in H₂P05, which is an aobhipratie species (acting both as acid and bases and its pH is found using appronilation, ph= pkai tp kaz = 13076.70 -14.00) At 100mL: Po (98) + H₂ON HP05łaq) + Du-lag), kb, = 104 104 Let at equilibrium: 2.4x25 kas 1067 (using My = Mr =0.48-ng Since kby is very small, do we i 1573 Then, Kn? (21) ~ a12 UTO48-) CH3 0+3 = 10% COH) Die = Jkb,048 = 1.55x1M= CÓW-] = 6.443x15" M, pH = -log (6.442816) = 110.19 () (0 - 32 10 = kx = CHP] [Hot] = 02-2.4(.05)-0.057) Mb. - - nn assume 7 <0.48,40 66.48-) 0.48.

Hence before addition of KOH pH is 0.49,

then at 25 ml of KOH addition pH is 1.30,

at 50mL pH is 4.00,

at 75 mL pH is 6.70 and

at 100 mL pH is 10.19.

Comment in case of any doubt.