Question

9 PM Sun Nov 17 1) Suppose a 0.33m aqueous solution of CH3 Pou) is prepared. Calculate the eluilibrium Properties of Phosphoric acid pka, = 2.16 pha 2 = 7.21 pk az = 12.32

Answer 1

pKa1 = 2.16 => Ka1 = 10-2.16 = 0.006918 pKa2 = 7.21=> Ka2 = 10-7.21 = 6.17 x 10-8 pka3 = 12.32 => Ka3 = 10-12.32 = 4.79 x 10-13

The dissociation of Phosphoric acid (triprotic acid) is given by H2PO4 H + Initial 0.33 H2PO4 Change + X + x Equilibrium 0.33 - X , _[H* ][H2PO4). X.X C =0.006918 saa [H2PO4] 0.400- x >x? +0.006918x -0.002283 = 0 By solving, x = 0.044447 [H*] =0.044447M [H2PO4 ]=0.044447M [H2PO2]=0.33 - x=0.33 -0.044447 = 0.285553 M = 0.286M

H,PO, will further dissociates into Ht and HPO..2 Let y be the change in the conc. of all species at equilibrium. H+ + HPO, -2 H2PO4 0.044447 Initial 0.044447 Change - y + y +y Equilibrium 0.044447 - y 0.044447 + y K _ [H*][HPO,2] (0.044447+ y) ).V = 6.17x10-8 [H,P0,-] 0.044447 – y Since y is very small, 0.044447 - y = 0.044447 and neglet y? 0.044447 y = 6.17x10-8 0.044447 :: y=6.17x10- (insignificant) [HPO, ?]= y = 6.17x10*M [H"]=0.044447 + y = 0.044447 +(6.17x10-* )= 0.044447 = 4.44x10²M [H,PO, 1=0.044447 – y=0.044447-(6.17x10-8)=0.044447 = 4.44 x10-²M

HPO4-2 will further dissociates into H+ and PO2-3 Let z be the change in the conc. of all species at equilibrium. Каз HPO4-2 6.17x10-8 H+ + 0.044447 P-3 Initial Change -Z + z +z Equilibrium 6.17x10-8 – Z 0.044447 + [H+][PO, ']_(0.044447 +-).- _ 1 7910-13 [HPO,?] (6.17x10-)-- Neglet z’ and assume (6.17x10-)-:-(6.2 x10-8) 0.0444472 -4 79x10-13 6.17 x 10-8 :::=6.64 x10-19 (insignificant) [H"]=0.044447 + = = 0.044447 +(6.64 x10-19)=0.044447 [HP0,-?]=(6.17x10$)--=(6.17x10-8)–(6.64x10-1)=(6.17x10-)M [PO, ']=-=5.99x10-1 M pH =-log[H+]=-log(0.044447)=1.35 [OH"]=- =2.25x10-13 M 0.044447 [H,PO,]=0.286 M. [H,PO, ]=4.44x10-2 M [HPO,?]=6.17x10-SM [PO, °]=6.64x10-1 M [H"]=4.44x10 PM [OH-]= 2.25x10-13 M. pH =1.35