Question

Part A Calculate the H,O of the following polyprotic acid solution: 0.390 MH,PO, Express your answer using two significant figures. O ACCO ? [H,01 = Submit Request Answer Part B Calculate the pH of this solution Express your answer using one decimal place. Val AEO ? pH = Submit Request Answer

Answer 1

Given that conc. of phosphoric acid = 0.370 M The phosphoric acid has the following ionizations H3PO4 + H2O = H2PO4 + H30+ Kan = 7.5 x 10-3 H2PO4- + H2O = HPO4-2 + H30+ Kaz= 6.2 x 10-8 HPO4-2 + H2O = PO2-3 + H2O+ Ka3 = 4.8 x 10-13

The dissociation of Phosphoric acid (triprotic acid) is given by H2PO4 + H2O - H0+ + Initial 0.390 H2PO4 Change -X + X + x Equilibrium 0.390 - X x _[H2O* ][H2P04) x.x -= 7.5x10- = 0.0075 [H2PO4] 0.390 - x = x² +0.0075x – 0.002925 = 0 By solving, x=0.050463 [H2O*]=0.050463M [H2PO4 ]=0.050463M Ka2 and Ka3 are much smaller. So, ignore the further ionization. [H3O+] = 0.050463 M = 0.050 M Part-A: [H3O+] = 0.050 M Part-B: pH = -log(0.050) = 1.30

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Given that pka1 of oxalic acid = 1.23 Ka1 = 10-1.23 = 0.058884 Ka2 = 10-4.19 = 6.457x 10-5 The dissociation of diprotic acid is given by H20204 - 0.370 H+ 0 + HC,04 0 Initial Change Equilibrium 0.370 - X H*][HC2041_ x.x =0.058884 [H.C,0,] 0.370- x x? +0.058884x – 0.021787 = 0 By solving x = 0.12107 [H+]=0.12107M [HC,0,-]=0.12107M

HC,04 will further dissociates into H+ and C,0,2 Let y be the change in the conc. of all species at equilibrium. CC HC2O4 0.12107 + + 0.12107 Initial Change - y + y Equilibrium 0.12107-y 0.12107+ y [+][0,0,?]_(0.12107 + y).Y –6 457x10-5 * a2 + [HC,0,-] 0.12107 – y Since y is very small,0.12107- y = 0.12107 and neglet y? L=6.457 x10- 0.12107 :: y =6.457x10- [H*]=0.12107+y=0.12107 +(6.457x10-4)=0.12107 [H"]=0.12107M pH =-log[H* ]=-log(0.12107)=0.92 PH=0.92 Part -C: [H,0*]=0.12M Part-D: PH=0.92