____________________________________________________
Answer:
Your answer would be a). 2.0 × 10-9
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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
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<em>-Julie</em>
Ag203 (Nh4)2co3 solution is Added to 79.8mL
The initial ph of the 0. 10M ch3cooh (aq) solution is -0.64.
The salt sodium acetate is present at the equivalency point. It is a salt of a strong base and a weak acid. The expression "its pH" provides the information.
pH = 7 + 0.5p + 0.5logC
where C is the salt concentration. Monobasic acid is Acetic acid here.
25 ml of 0.10 M NaOH solution are needed for the titration of 25 ml of 0.10 M acetic acid.
Total volume = 25 + 25 = 50ml
The salt concentration = C = 0.10× = 0.05
The pH for weak acid is given by,
pH =
(p
+ logC)
pH =
(1.8× + log0.05)
pH =
[1.8× + (-1.30)]
pH =
(1.8× -1.30)
pH = -0.64
Learn more about pH here;
brainly.com/question/15192160
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