Calculate the pH of the solution made by adding 0.50mol of HOBr and 0.30mol of KOBr to 1.00L of water.?

Calculate the pH of the solution made by adding 0.50mol of HOBr and 0.30mol of KOBr to 1.00L of water. The value of Ka for HOBr is 2.0x10^-9.

Answer:
This is a buffer problem.

ICE table should be used

Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]

0.50 mol in 1.00 L = 0.50 M HOBr
0.30 mol in 1.00 L = 0.30 M KOBr (or OBr-)

HOBr + H2O <-> H3O+ + OBr-
0.50 0 0.30
-x +x +x
(0.50-x) x (0.30+x)

Assume x is small comparing to 0.30 and 0.50

2.0 x 10^-9 = x (0.30) / 0.50)

[H3O+] = x = 3.33 x 10^-9
pH = 8.48
pH= pKa + log [(Base) KOBr]/ [(Acid) HOBr]

pKa = -log Ka

[KOBr] = mole / L
[HOBr] = mole/ L

you will get the answer.

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