Question

Phosphoric acid, H,PO (aq), is a triprotic acid, meaning that one molecule of the acid has three acidic protons. Estimate the pH and the concentrations of all species in a 0.300 M phosphoric acid solution. pkalpK42 2.16 7.21 PK 23 12.32 H,PO4) = [H,P0] = M [OH") = HPO ) = M pH = | PO]=

Answer 1

- To find Kai, Kaz Kaz: we know pk = -log (ka) Given pral = 2.16 = 2.16 = -log Ka, as Logka, - 2.16 - Kas Antilog (- 2116) Antilog (-2.16) = 0.0069 (Kai = 6.9x103) ar 6.9x103 Given pra2 = 7:21 - 7.21 = -log ka, Ka = Antilog (-7.21) = 6.2x108 Reag - 6.2 X16*7 Given Prag = 12:32 Raz = Antilong (-12-32) = 4.8 x 103 Kaz = 4.8x10% 12 .3 Stepwise dissociation of H₂ Poz is given as: H₃ Poy (98) ² Hapoy + H 2 ; Ka, = 6.9x103 - 0 (99)_ 19%) 6 H₂ pou (199) ? Hoy² + 4 + ; Kq = 6.2x108 Pers) (98) Apoy (92) ? Poy3 + Ht ; Ką = 4.88103 € 199) 082

Hoy lesing H3 Poy 2 4 + Initial ,0.300 0 Change ax to equilibrium 0.300-2 a Ka, is given by : Ka, = [47] [H₂poy [H₂ poy] → 6.9x10²= din 0 the a rin 0.300- Xlon 22= 60-300-) (6-9x103) x² = 2.07810 ² - 6,9x10²n er laget fuq x ton – 2:07x17 250 ( - - - - y = -6.4xło?+ 11619813332_48147-2:0747037 2x1 -619x10+ J 47.61 70668.28 x 103 X = x=-6.98103 + 747-614póft 8280 xt66 - 2 x = -69x703 + N8327-61xTor -6.9x103 + 91-25 x1o3 2

-6.9x103 + 91.25x163 | OR -6.9x103_91-25 x 7o3 = - x= 84.358103 2 from here we will get negative - value ofx. But value of re can't be negative so this can be neglected. X = 42.17X102 [se = 0.042] At equilibrium [Hz Poy) - 0.300 -x2 = 0.300 – 0.042 = 0.258 M - :- (Hz Poy] = 0.258 M OR 25-8x16M 902 and [H2804]=[++] = x . (H₂Poy) = 0.042 Mr Similaky (ht] = 0.042 M ✓ hsing eq - H₂ Pon ² HPott 1. Initial 0.042 o 0.042 -X th change th At equilibjум боули О.o42 -и и Ka = (hPoy? ] [ht] - [H, Pon]

- → 6.1X10% = 0.042+n) Cod 10.042-) Let us assume a is very small an Compased to e can be neglected 0.042-un 0.042 4 0.042tna o.o42 0.042 so - 7 6.2 x 108 = 0.042 Xx 0.02 1 * = 6-28109 n At equilibrium (H₂Poi] = 0.042-2 . (H, pou J = 0.042 – 6.2x10 © very very - f[, Pou) = 0.042 M or 4.2xjö?m 0.042 M small (HPou?) = x . HPoy = 6.2X108m ~ [let) = 0.042 + x : [frt] = 0.042 + 6.2*100m ñ 0.042m (H+) = 0.042_M

using eq ② : неоу Initial 6.28108 - н+ + fo, 3 0.042 o ta tx Change an Equilibrium 6.2x lo-n 0.042th n (ht] (Poy3] (hPoyz => 48840'3- (0.042+1) (n) XD - 6.2x108-11 Assume neck 6.2X100, son can be neglected = 4,8x1013 0.042.2 612xto 8 > m= 4.8 X 103 x 6:2x168 8042 x = 997644574 708-578 TL -0.042 w Tube x15 19 :-* = 7.08x169

At equilibrium : (H2Poi = 6.2 X108 - Et pow] = 6.2X10?– 7.08 x 70' 9 ñ 6-2X1o8 M very morning Mall - Etapat = c2x10* Ponuda 70pxjöl9 | [Poy SR = x (ht] = 0.042+ x = 0.042+7.0881019 = 0.042M Chet = 0.042 OR 4.2 X 102 M we know PH = -log [ut] pH = -log [4.2 x 10² ] [loga xb= loga + hoy b] PH - [ log 4.2 + logróz ] [ log ab = bloga] +- 0.623 -2 lag (0) [tag 10-1] 2 - (0.623-2) = -(–1:31) = +1:37 P= 1.371

Jonization Constant of water is given by : [ht] [or] = Kw or (or] = ko v we know Kw = 1x1014 (H+] = 0.042 = [OH) = 1x10th = 23.8 x1014 z ox 2138* 16'3 M - ("]e_2:38KJ013 m/v '= 23.8 x 1019 0.012 Ans: Hapoy1 = 0.258 M or 25.8810² M ✓ [ha Poj] = 0.042 M as 4.2810² m v Chpo ] = 6.28108 M V [Poys] = 7.08x1619 Mv [H+) - 0.042 or 4.2 X16XMV (OH) = 2.38 x1573 M V PH = 1:37 -