Question

LABORATORY 7 HYDROLYSIS OF SALTS AND PH OF BUFFER SOLUTIONS 3. How should the pH of a 0.1 M solution of NaC,H,O, compare with that of 0.1 M solution of KC,H,O,? Explain.

Answer 1

Ionization of CH,COONa CH,COONa(aq) →→ CH,COO (aq) + Nat(aq) Ionization of CH,COOK: CH,COOK (aq) – CH3COO-(aq) + K+ (aq) Since Imol of CH,COONa form simol of CH,COO = [CH,COONa]= [CH,COOK] = 0.10 M Equation representing the ionization of CH,COO is as follows CH,000- + H,0 —— CH3COOH + OH- Since the pH of acetate salt is due to the hydrolysis of CH3COO- Since the concentration of CH,COO is same in both the salts , pH of both of the salts will be the same - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Ionization of CH3COONa CH,COONa (aq) → CH,000 (aq) + Na+ (aq) Since Imol of CH COONa form simol of CH,COO = [CH,COONa] = 0.10 M Equation representing the ionization of CH,COO is as follows CH,COO + H2O —— CH,COOH + OH- Acidionization constant K, of CH2COOH = 1.8x10- CH,COO is the conjugate base of the acid CH COOH Baseionization constant K of CH, * 1.0x10-14 1.8x10= = 5.6x10-20 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - CH3COO + H20 —— CH,COOH + OH Initial concentration (M): 0.10 0 0 Change(M) -X + x + x Equilibrium concentration (M): 0.10- x [CH,COOH]OH-] Baseionization constant, K = 5 [CH,000 ] 5.6x10-10 (x)(x) 0.10-X x = 7.5x10M From the equilibrium table, x = [OH-] = 7.5x10M [,0'] = Core 2 3*10* = 13410 pH=-log[H20+] =-log(1.3x10") = 8.88