Question

Calculate the pH of a solution containing 50 mg/l of each of the following weak acids or salts of weak acids:

a) carbonic acid

b) sodium acetate

c) sodium hypochlorite

d) phosphoric acid

Answer 1

50 mg /L carbonic acid-50 "Sx 1s 1mole H2Co 62.03g L 10' mg -8.06 x104 M This is a weak acid. It will dissociate partially giving H ions. Let change in concentration be x. Sethup the reaction table as follows K, 2.51 × 10-4 Initial Change Final 8.06x 10-, M (8.06x10*-x) x x H2COs 2.51x104Xx 8.06×10- -x On solving,x0.000341 So[H]-0.000341 M pH- -log [H] - log (0.000341)-3.47 b) mg 1g lmole sodium acetate 50 mg/L sodium acetate 50 x L 10 mg 82.03 g 6.10x10M This is asalt. It will dissociate fully giving acetate ions. Acetate ions are conjugate bases of acetic acid. -14 10 K. (acetic acid) = 1.74x10-5-75-10 Let change in concentration be x. Setup the reaction table as follows CH3COO-H20 6.10×10-4 M CH3COOH + OH Initial Change Final 6.10x104-x CH COO X. X 5.75x10- 6.10x10-x -10 On solving, x = 5.92 x 10 So OH-5.92x10 poH--log-OH-]--log (5.92x 10-7)-623 pH 14 pOH 14-6.23- 7.77 Sodium hypochlorite is also a salt. It dissociates completely giving hypochlorite ions. Hypochlorite ions react with water giving OH ion:s. 14 10 K. (HcIo) 2.95x10 .rā = 3.39x10-7 Follow the same procedure as in part b to obtain the pH

mg, 1g 1 mole phosphoric acid Concentration of H.PO, 50-ー×ー.e- L 10 mg = 97.9% = 5.10 x 10-4 pK of H,PO 2.16 2.16 6.92 x103 ICE table constructed for the ionization of H PO4 is as follows; HyPO,(a) H20()HPO,(aq) + H,O' (aq) 5.10x10-4 I(M): C(M) E(M): 0 5.10x10-4-x LH,POTIH, 1st acidionization constant, K - H3PO 6.92x10 5.10x10-4-x Solve for x and use the concentration of H in the next two equations H,PO, (aq) + H2O(l)--HPO42- ) Ka,-6. 17 ×10-8 HPO,'" (aq) + H2O(1)ミ- -PO43-(aq) + Hợ (aq) K,,-4.79 x 10-13 Repeat the same calculation and calculate final pH Final pH will be around 3.31 (aq) + H,O*(aq