Question

Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.133 M barbituric acid (K (a) no KOH added 1 244 - 9.8x105) with 0.133 M KOH. (Assume K - 1.01x10-44) (b) 20. mL of KOH solution added 04.01 (c) 39 mL of KOH solution added 5.6 (d) 40. mL of KOH solution added (e) 41 mL of KOH solution added 11.09

Answer 1

Cu Ma Osona (ap+ Hy (8) = 40*Can ) + Cung Sony Ka= [14.30 t] [H2O2N2] Се «0,9) [CulHGOzae] [Hot ] [author] In ihal concentra has 0.133 change (m) an i n Equilibrium (m) 0.100- ka= 9.8 x 10's (given) assural Ca (2,133 - x 4 -133 x = 0.133 M Z ka = [1to+] (culkaa Ná) [culta Oz N2] 9.8x10 s = [n] Cu ] = 42 [0.133-4) [0.133) x² = 1.3034 x16 5 = 0.13034 x 10 4 u= 0.36103 x 10-&, 01003 6103

[itzo+] = otu = 0.0036 m - pho-log CH') = -logo [0.0036 ? 2.444

(6) The moles of ot added are given by; mol (olt-) = mx v = (01133m x 0.0204) = 0,00266 m) The inhal moles of barobihoic and mol HA - MXV - C0.133 x 0.040L) -0.005 32 If the added bydanude lon reacts completely with an equal number of moles of HA , forming an equal number of moles of A in the proup. The moles of lone are given by : mol HA. 0.00532 -0.00266 = 0.00266 mol Ara 0.002 66 mol Total volume is : HomL +20mL = 60ml = 0.0600L ihal concentratas : = 0.0443 in CHA=0.00266 0.0600 [A") - 00443 On substhiting: [Hot][A CHA] ] = n (0.044 3 th) n (0.0443 th). x x00HH3 (0.0443-2) 0.0443 =9.8x105

pH = -log (9.8x10-5) = 4.009 :4.01 (c) The modes of COH) added: mol OH = mxva 0.133 *0.039 L 20.005187 Inihal moles: mol HA = MXV = (0.13 3 x 0. 040) = 0.00532 mol. Added Ot in reacts completely with ha, formy an equal no. of moles of A- in the proces, mo IHA = 0.00532 -0.005187 = 0.000133 ega mol A = 0.0039 Total volume 40 +39 = 7amL = 0.0790L Inital ancontraha : HA = 0,000133 0.00168 0.0990 = 0.0039 0.079 Hot] [ A THAJ 0.0494 = u (0.0494+4) (0.00168-4) Subshhity: on x 29.40S = 9.8 xios = 0.033 x107

pH = -log (0.0331104) asing (d). The moles of olt : mol OH-. Mive 01133 x 0.040 - 0.00532 mol HA = myv = 0.133 x 0,040 = 0.00532 mal At equivalence point where the moles of base added equal the moles of and present mol A = 0.00532 TVE 40 mL + HomL = 80 mL 20 108L A = 0.00532 = 0.0665 m 0.08 Reochon will be : Acaq) + 40 (1) kn = [HA]COH) - kw TAJ Ka HA cag) + OH Cor) a 1.000 19. 9.8to" = 1.08x10'' -lo substhay: THAOH TAJ : (M) (2) x2 = todo (0.0665 -4) 0.0665 = lodrio lo

22. 0.06783 x 10 10 nc - 0.2604x10-5 plom) - - log (0.2604x10") • 5.58 pH = 14-5.58 = 8.42 (e) (comL +41ML) Total vodune = 0.081 L [OH ) - 0.0 mol Colt ) = 0.133 x00u1 = 0.005453 mol HA · 01133 x 0.040 -0.00532 mol OH - 01005 #83-0.00532-0.000133 TOH-] = 0.001642 0.000133 0.0810 p[on] = -log (ort] = 2.485 pH = 11:21