The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?

a.2.0 × 10-9

b.1.1 × 10-9

c.6.0 × 10-5

d.3.3 × 10-5

e.3.0 × 104

Your answer would be a). 2.0 × 10-9

Work:

In your question the "ph" of a 0.55 m aqueous solution of hypobromous acid temperature is at 25 degrees C, and it's "ph" is 4.48.

You would use the ph (4.48) to find the ka for "hbro"

[H+]

=

10^-4.48

=

3.31 x 10^-5 M

=

[BrO-]

or: [H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-]

Then you would find ka:

(3.31 x 10^-5)^2/0.55 =2 x 10^-9

-Julie

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.