Question

A solution is 0.125 M in propanoic acid (CH.CH,CO,H) and 8.33x10M in sodium propanoate (CH,CHCO,Na). What will be the pH after adding 0.0100 mol of NaOH to 1.20 L of this solution? Assume no change in the volume upon adding NaOH. K for propanoic acid = 1.3 x 10. 0.125 A) 6.79 B) 4.64 C) 5.06 D) 9.36 E) 4.78 Which statement is correct for the reaction represented below? 2 NOCI(g) → 2 NO(g) + Cl2(g) AH = 77.38 kJ/mol The reaction is: A) More information is needed to determine the reaction spontaneity. B) Spontaneous only at high temperatures. C) Spontaneous only at low temperatures. D) Not spontaneous at any temperature. E) Spontaneous at all temperatures.

Answer 1

1) Concentration of CH,CH,COOH, (CH,CH,COOH) = 0.125 M Concentration of CH, CH,C00,[CH,CH,000 ]=8.33x10^M Concentration of base(NaOH) added = 1.20 L 0.0100 mol-0.00833 M + CH, CH,COO+ H,0 ICE table: CH,CH,COOH + OH- I(M): 0.125 0.00833 C(M): -0.00833 -0.00833 E(M): 0.11667 0 8.33x102 + 0.00833 0.09163 PK, of CH,CH,COOH = -log K = -log(1.3x10-) = 4.89 According to Henderson equation, CH,CH,000) pH = pK, + logo [CH,CH,COOH] 0.09163 - 4.78 = 4.89+ log o 11667