3. To prepare a buffer of pH 7.5 Given acid = HBrO conjugate base = BrO- pKa of
ID: 700113 • Letter: 3
Question
3. To prepare a buffer of pH 7.5
Given acid = HBrO
conjugate base = BrO-
pKa of acid = 8.64
Using Hendersen-Haselbalck equation,
pH = pKa + log(base/acid)
7.5 = 8.64 + log(BrO-/HBrO)
0.072(HBrO) = (BrO-)
Now,
(HBrO) + (BrO-) = 0.1 M x 75 ml = 7.5 mmol
From above,
(HBrO) + 0.072(HBrO) = 7.5 mmol
(HBrO) = 7.5 mmol/1.072 = 7.0 mmol
(BrO-) = 7.5 - 7.0 = 0.5 mmol
Volume of NaOH to be added = 0.5 mmol/0.5 M = 1 ml
4. Volume of NaOH to reach equivalence point = 0.1 M x 75 ml/0.5 ml = 15 ml
[BrO-] formed = 0.1 M x 75 ml/(75 + 15) ml = 0.0833 M
Hydrolysis of BrO-,
BrO- + H2O <==> HBrO + OH-
let x amount reacted
Kb = [HBrO][OH-]/[BrO-]
pKb = 14 - pKa = 14 - 8.64 = 5.36
pKb = -logKb
Kb = 4.36 x 10^-6
4.36 x 10^-6 = x^2/0.0833
x = [OH-] = 6.03 x 10^-4 M
pOH = -log[OH-] = 3.22
pH = 14 - pOH = 10.78
Explanation / Answer
AT&T; LTE 4:51 PM Done 12 of 12 3. 18 pts) What volume of the 0.5 M NaOH(aq) solution would you need to add to the 75.0 ml of 0.1 M acid solution to obtain the buffer with a ph of 7.5 727 HBO 20-75 ey ex?) 4.8 pts) If you continued to add as M NaO-aal toyour acid sektion unti you reached the equivalence point, what would the ph be? 6.02x10"/mol 4 ]
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