16.22 A 100.0 mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO.
a. What is the initial pH of this solution?
b. What is the pH after addition of 150.0 mg of HBr?
c. What is the pH after addition of 85.0 mg of NaOH?
-8-
HClO K
= 2.9x10
a
SOLUTION:
It’s a buffer, H-H will work.
-8
pK
=-log(K
)=-log(2.9x10
)=7.54
a
a
(
)
HBr is a strong acid, it dilutes an equivalent amount of the best base available: ClO-
Since I’m using H-H, I can do everything in moles.
0.150 M x 0.100 L = 0.015 moles
0.175 M x0.100 L = 0.0175 moles
-
→
HBr +
ClO
HClO
+ Br-
I
0.001854 mol
0.0150 moles
0.0175 moles
0
Neutralized -0.001854
-0.001854
+0.001854
+0.001854
Left
0
0.013146
0.019354
0.019354
(
)
NaOH does the opposite: neutralize the equivalent amount of the best available acid HClO
→
OH- +
HClO
ClO-
+ H
O
2