Question

2. The pH of a 1.79 M solution of a weak base B is measured to be 11.64

Determine K_b of the base


Determine K_a of the weak bases's conjugate acid, HB⁺


Determine [H⁺] in a 2.00 M solution of the chloride salt, HBCl
M

Determine the pH of a 2.00 M solution of the chloride salt, HBCl

Answer 1

(9) given that fИ се деаk Rege Ball. ( 4 ean Conc So, we can write =) - B+H₂O 2 BHO ton - Intial C eres - - final c-cx - cx. cx » Kolo = [But] [or] [B] Ку - С. С 22 c-c4 ecia) gnore Ignore & Wort. I = we get Ky = CR? 17 x = I thin 491972) .X et given that pH= 11.64 = -log(n+] = 11.64 =) H+ = tolhcy = 2.29 xlól?

(C) given that am Sulen of HBcl Salt, Иlot 0 — «В» и со- Gurial, 2m - - final - 2M 2M = (But = 2m) And Bli will dissociate as - But Bt HO 9" @ gurial final c- cq cx cx = Ke= cekey = 222 1 C-ca el1-8) Here Ka from part (b) we know that Ka = 9.394x100 C = Cncof But' = pm =) ( = 939yris". = 14.6770 YH50 [X= P. 16 X105

=) [ht] flom ep ③ , we get → [ht) = CX = 2x 2.16 x175 (41) = 4.33 x1ös muy (d) Now, pH = -log[rit] 3-10 [423x105] py = -(-4.36) = 4.36 [ PM = 4:36 (As.

According to Sonic poduct of water =) [111] [o1). Loty 1 Coy] - igby - = 0.0044 [117] P29x1712 from qh o 106) - CK = 0.0044 *= panya orang [s carried - 0.0044 - chest 179 K = 0.000438 30; forn en ③ we get =) Ky z cx'= 1.19 x(0.002438) Ko= 1.064X105 muss. we have to Calculate ka of the Bn . We know that a kaxkg = kw Ky =) kq = kw = 104 Ka = 9.394 X 1000 me 1.064X165