Question

(b) Calculate the percent ionization of 0.00640 M hypobromous acid in a solution containing 0.0790 M sodium hypobromite.

(a) Calculate the percent ionization of 0.00640 M hypobromous acid (Ka = 2.5e-09) % ionization = (b) Calculate the percent ionization of 0.00640 M hypobromous acid in a solution containing 0.0790 M sodium hypobromite. % ionization =

Answer 1

Blution Given or data .. 1 hypobromous acid Ka 20.00 6uo M a 2.5 410-09 Brow -> Bro +H+ Initial 0.006uo . x x Fruilibrium 0-c06@-

[ro] (the ot] Sasklon 6.00CRO-X U = 0.00640 x Svaly CHOO 640 - 2.5810- = gel - 0.0064.0 . 1.6460-1l M. * = 1.6xre "M = 2t] = 1.6410-UM foot Xloo i gonization & THOBI 3.0/10-6 Xloo 0.00 600 = 0.046%

(13) typobromous acid obromite = 0.00640 M = 0.0790 M 4-toBr + H20 - Initial oroobuo change - Bro 0.079 +H30- o 0.079 +2 -0.006uce Equilibriune ta - (Bro-) (teet) (HORY) 254601? = 10.019+x) xx HOBY 0.006402 ouis Vay 00 64, Sumi Se 0.00602 = 0.079 : 0.0079 + x 215x10 -4 = 0.0792 . 0400 64 22.2.02x10-10 M (Hyot) = goodx10-10 M -

% ionization =1t307 HoBryxion a 3-15 X10 x100 3. 15 X10