Question

A 25.00 mL solution of 0.03220 M Na2CO3 is titrated with 0.02510 M Ba(NO3)2 . Calculate pBa2+ following the addition of the given volumes of Ba(NO3)2 . The ?sp for BaCO3 is 5.0×10−9 . 13.90 mLpBa2+= ?epBa2+= 37.40 mLpBa2+=

Answer 1

Reaction of Narcouf Ba(NO3)2 Na co (aq) + Ba(NO3) a) Bacon (6) a + 2NONO (ar) at equivalence point all racon will be converted to Bacoz. Art canivalence according to stoichiometric ratio of this reaction moles of Nazcoa = Moles of Ba(NO3ler 2) Nagozwanson & Mba (Non) & Vep 0903220M X 25:0 = 0·02510 MM Xvep / Vep: Volume of BONogle 0.0 3220 x 25 and needed to Seah eanirdre = 32.00 point MP Molarity when, 13.90m Ba(Node is added Ve volume] then in solution Naces will remain exeen. All the Ba(NO), will getprecipitate as Bacog. So, in solution Batt mill- come due to dissociation of Bacoz. Now ex con (cod. Narcos Non con Manon eodread wonlon & Ba(NO3)2 (0.03240, -223)60.07510 X 13.9) & 18.9 Vorancez n (254139) = 0.0117 in vep - 0.02510 There (Noh

wow let in presence of excen cchr, the solubility of Bacon in s mohl. Bacon dissociates ar BOCOR (s) Bart (az) + 0 0. Smilia 0.0117 afterdi - - S (0.017+5). wow, expression of solubility Product of b4 с. : Rohr Barrx C coz CooXPO-9. S X (0.0114+s) kop in [cz comment on SX 0.0117 2 50x70-7 & Su, 0.0117 ts .: S. si ox?ons 427 xx Fmo! O 0117 E 427x10 m in CBalt=4277157 eur up Balt: – Log Cart – –log (4.27 807) (10 . # = 6.37 t 13.90 Ba(NO3)2 Bart. 6:37 At canivalence point there aull no be any excon clar det solubility at this time in s'more Chat as molk & C cogan & Smoke . kop & CB alt x C con? G x 7 = x =) SoX10-9 Se (510x1891 707x10 mol/L

#: at eanävalence point Baltu 7.07 x 10 more & Bart og CBart & log (7.0 780-5) = gols # When 37 yow of Ba(NO3), is added then, excess Bart acill remains in solution. MBA No3) Balon Mwasas acon he exon (Barthur VBANO + Vascon (0.025108 37 4. (o. orrores) (37 4 + 28 - 2.148103m .: pbaat olog Cha247 om slag, (2.04 810 ) - 2067 V 10