Question

For the diprotic weak acid H2A, Ka1 = 2.7 × 10-5 and Ka2 = 5.2 × 10-7. What is the pH of a 0.0550 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Answer 1

Tofind the pH of a 0.0550 M soluti on of H2A Given that K27x10 al K25.2x107 It will di ssoci ates H2A HAH' K, = 2.7x10 HA A2H* K, =5.2x10 If x moles of H2A dissociate, then x moles of HA and H'are produced and the concentration of H2A is reduced by x [HJA] 0.0550-x E HA-]x Ht = X Equilibrium expression HA K HA al 2.7x 10 0.0550- x since 0.0550 -x is very small 0.0550 - x 0.0550 x2 2.7x 10 x0.0550 x 2.7x 10 x0.0550 x 0.001218M Since x represents equilibrium concentration of hydrogen cations Hx 0.001218M To find the pH pHlog H =-log (0.001218) pH 2.91 Hence the pH 2.91

Tofindthe[H2A H2A 0.0550-x H2A 0.0550-0.001218 = 0.0537M Tofindthe A2 For the secondreaction HA A2H K,2=5.2x107 Equilibrium expres sion HA [A2- K2 A K 2-5.2x10M A Hence the pH 2.91 H2A 0.0537M A2=5.2x10- M