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A 60.0 ml simple of 0.100 m nabob is t if rated wi
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Answer #1

12) a ) before titration only NaOBr present in the solution.

NaOBr is salt of weak acid and strong base

for sucg salts

pH = 1/2 (pKw + pKa + log C)

pKw = 14

pKa of HOBr = 8.7

pH = 1/2 (14 + 8.7 + log 0.1)

pH = 1/2 (21.7)

pH = 10.85

b) milli moles of NaOBr = 60 x 0.1 = 6.0

at equivalent point all NaOBr becomes HOBr

6.0 milli moles of HCl must be added

6.0 = V x 0.3

V = 20

total volume = 20 + 60 = 80

[HOCl] = 6.0 / 80 =0.075 M

HOCl is weak acid for weak acids

pH = 1/2 (pKa - logC)

pH = 1/2 (8.7 - log 0.075)

pH = 3.79

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