2 points 3 points 3 points 3 points A buffer was prepared by dissolving 0.215 mo
ID: 783498 • Letter: 2
Question
A buffer was prepared by dissolving 0.215 moles of hypobromous acid along with 0.013 moles of its conjugate base (sodium hypobromite) in water to a total volume of 500 mL. What is its pH? (report to 2 digits) The pKa of HOBr is 8.63.
A buffer was prepared by dissolving 0.215 moles of hypobromous acid along with 0.013 moles of its conjugate base (sodium hypobromite) in water to a total volume of 500 mL. The pKa of HOBr is 8.63.
What is its pH if 0.023 moles of NaOH is added to the solution assuming no change in volume? (report to 2 digits)
Why does a 1 x 10-8 M solution of strong base KOH NOT have a pH=6. When we know a strong base is 100% dissoiated so the we can calculate the pOH as pOH = -log[1x10-8] = 8. Then pH = 14-pOH = 6.
That is the correct pH, and there is no need to explain it.
The autoionization of water was not considered in the simple equation, and it is important near neutrality.
The solution is a buffer and the Henderson-Hasselbalch equation was not used.
The effect of ionic strength, and hence activity was not considered.
10.22
A < B < C
C < B < A
B < C < A
C < A < B
Explanation / Answer
1)C < A < B
2)3.27
3)4-aminobenzenesulfonic acid
since pka=3.23
4)The acid's pKa is close to that of the desired buffer pH.
5)pH=pka+log(salt/acid)
=8.63+log(0.013/0.215)
=7.41
6)pH=pka+log(salt/acid)
=8.63+log((0.013+0.0230)/(0.215-0.023))
=7.9
7)The autoionization of water was not considered in the simple equation, and it is important near neutrality.
8)pH=pka+log(salt/acid)
=9.38 +log(0.021/0.003)
=10.22
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