Question

A 25.00 mL solution of 0.03060 M Na,Co, is titrated with 0.02550 M Ba(NO3), Calculate pBa?' following the addition of the given volumes of Ba(NO3), The Kep for Baco, is 5.0 x 10-. 14.90 mL pBa²+ = V. pBa²+ = 37.60 mL pBa2+ =

Answer 1

Now ва сода Bat + со nn Ksp = [Ba²+] [co 22-] KSP 2 n2 na ſ Kop & √5x109 na 7.07x10-5 moles. This gives the mole of Ba²+ and CO - When 14.90 mL of Ba(NO3), is added in the 25ml Na2CO3 solution, Total volume of solution a 14.90 + 25 z 39.90 mL from Ksp data, moles of Bq2t - 7.07x10 moles Conc, a n e no od moles vol. [Ba²+] = 7,0 78100 mol = 1.81x10 3 M 39.90 x 10-3 L pat -log[Ba2+ ] z log (1081x103) I Bar 2.7427 At equivalence point, total volume 225 +30 245 mL - 7007810 mol 20.00l57M [Bat] Ž 7463102 pbåt - log[Ba") = -log(0.00157) [ pat = 2.80

- At 37.60 mL titratt addition, Total volume 2 (87.60+25 mL 2 62.60 mL (Baht) = 2 a 7.07x10 moles . 62.60 x 10-3 L = 0.00113 M Bat - log [Ba²+] = -log [o-00113] [ PBa?t a 209469 |