Question

al 11. Calculate the pH of a 100.0 ml buffer solution with a composition of 0.100 mol L'HPO and 0.162 mol L-! H.PO,(H3PO, has K. - 7.5*10-); Ke-6.2x10; K3 = 4.2*10-13) 0.1000 LX0-100 M:0.0100 molt pH = 1.79 11 H₂PO4-3=0.0617M HPO42-+ Kan.6.2x10 3.83x102 012- 720 H₃PO4 + OH 0.0162 12. Calculate the pH of a buffer solution composed of 30.0 mL of 0.100 mol L of HC,H,O, and 0.100 mol L NaC,H,O, after addition of 2.00 mL of 1.00 mol L of NaOH. (HC,H,O, has pK, -2.39). 0.0 300 LXO. 100 M = 0.003mola base added 0.00300 moyo 3mmel, a :0.00 100ml 0.00200 L x 1.OOM=0.00200 ml b=0.00500-o!L 13. Calculate the pH at the equivalence point of a titration of 25.00 mL of 0.75 mol L codeine (CH2,NOs) with 0.35 mol L-'HCI. (C13H21 NO, has Kb = 1.6*10%) [HC1] =0.0536M C is He, NO3 + the H₂O* + Cig H2o NO₃- 2.00 2.00 I pH= 3.og DH=6.13

Number 11 and 13 the answers on the box are wrong

Answer 1

11)

pH = pKa2 + log [HPO42- / H2PO4-]

     = 7.21 + log [0.100 / 0.162]

pH = 7.00

13)

mmoles of B = 25 x 0.75 = 18.75

At equivalence point :

mmoles of acid = mmoles of base

0.35 x V = 18.75

V = 53.57

volume of HCl = 53.47 mL

B   +     H2O     ------------->   BH+ +    OH-

At equivalence point : salt only remains.

salt concentration = 18.75 / 25 + 53.47 = 0.2386 M

pH = 7 - 1/2 (pKb + log C)

    = 7 - 1/2 (5.80 + log 0.239)

pH = 4.411