Question

Hey guys, I am having a lot of trouble with my homework. Could you please help? I thumbs up anyone who gives a thorough explanation and shows the work! I really need to understand these concepts!

1) Z- is a weak base. An 0.350 M aqueous solution of NaZ is prepared. The pH of the solution was 8.93 at 25.0°C. The Kb of Z- is: 2) The base-ionization constant of ethylamine (C2H5NH2) is 6.4 x 10-4 at 25.0 °C. The [H+] in a 1.6 × 10-2 M solution of ethylamine is M. 3) The acid-dissociation constants of phosphoric acid (H3PO4) are Ka = 7.5 x 10-3, Kaz = 6.2 x 10-8, and Kaz = 4.2 x 10-13 at 25.0°C. What is the pH of a 2.00 M aqueous solution of phosphoric acid?

Thanks guys!

Answer 1

7- (aq) + H2O(l) -> Hz (aq) + OH- pH 8.93 pOH = 14 - 8.92 5.07 = [OH- = 8.51x10-6 M Again [OH-] = kg x ¢ kb xc = [ok]² (8.51210-672 =) vo. 0.350 = 2.07*10-10 For ethyl amine (chy CH3NH2), Kb = 16.4x1024 Concentration ca 16*10-2M [OH-] J KG XC & J 6.4x104x16x182 3.2 x 103. Miris Now MX2/18 [H+][0] = 10-14 [H+] 10-14 3.2x103 = 3-125x102M

H₂PO4 (ag) H₂PO2 (aq) + H+ (as) Ka, = 7.5% 10 3 other dissociation constant (Kazandkaz) are can be neglected for the sake of simplicity. small and C = 2.00M so [H+ = kaxc 117.5*10-3 x 2.00) o. 1225M PH -log[ht = -log (0.1225) 0.9112