Question

What is the pH of a 0.1 molar sodium acetate(NaC2H302) solution? What is the concentration of each species present in the solution? (10 points) Ka(acetic acid) = 4.6 x 10-5 mole liter1 Kw = 1.0 x 10-14 mole2liter2

Answer 1

Ionization of NaC,H,O, is as follows: NaC,H,, (aq) = C,H,0, (aq) + Na+ (aq) Since im ol of NaC,H,O, forms 1mol of C,H,0,- [C,H,9,- ] = [NaC,H,] = 0.1 M ----------------- ----- Equation representing the ionization of C,H,O, is as follows C,H,0,+ H,0 — C,H,O, + OH Acidionization constant K, of C,H,O,= 4.6x10-mol/L C,H,O, is the conjugate acid of base of C,H,O, Base ionization constant Kof CH30; 1.0x10-14 mol/L 4.6x10 mol/L = 2.2x10-10 mol/L ----------------- ------------------- Draw the ICT table as follows: C,H,O, + H,O —— CH.0, + OH I(M): 0.1 0 0 C(M) - x +x +x E(M): 0.1-x + Base ionization constant, K. - CH.,](OH-] Baseionization constant, CH,02] Substitute the correspodning values in the above equation and calculate x. 22x10-40 mol/l- (*)(x) 0.1-x Assuming, 0.1-XN 0.1 2.2x10-1 mol/L= x = 0.22x10-10 (mol/L) x= 0.47x10mol/L ------- According to the equilibrium table, [C,H,O, ] = 0.1-x=0.1-0.47x10-* = 0.0999953 M [C,H,02]=x = 0.47x10-mol/L [OH-] = x = 0.47x109 molt ------ -------- ------ ------ ---- --- Concentration of hydronium ion, H, 1.0x10-14 0.47x10-5 = 2.1x109 - - - - - - - - - - - - - - Calculate the pH as follows: pH = -log[H,0*] = -log(2.1x10") = 8.68 Therefore the pH is 8.68),