Buffers Worksheet With Answers Page 4
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4Chem 1B
Dr. Abel
4
using one of the solid materials and one of the 1.00 M solutions (HCl or NaOH) so that
the concentration of the acid in the conjugate acid-base pair is 0.100 M. Give the mass
in grams and the volume of solution in mL.
As we learned in lab (and lecture) there are 2 ways to do this
a) Use NaH
PO
and NaOH:
2
4
2−
[HPO
]
4
7.00 = 7.20 + log
[H
PO
]
2
4
2−
2−
2−
[HPO
]
molesHPO
molesHPO
4
4
4
= 0.63096 =
=
−
−
0.100
[H
PO
]
molesH
PO
2
4
2
4
2−
molesHPO
= 0.063096
4
-
-
2-
H
PO
+
OH
→ H
O +
HPO
2
4
2
4
Start:
y
0.063096
0
React:
-0.063096
-0.063096
+0.063096
€
End:
0.100
0
0.063096
y-0.063096 = 0.100
-
y=0.163096 moles H
PO
2
4
NaOH: 0.063096 moles x L/1.00 moles x 1000mL/1L = 63 mL 1.00 M NaOH
NaH
PO
: 0.163096 moles x 119.98 g/mol = 20. g NaH
PO
2
4
2
4
Plus water to give 1 L total volume
b) Use Na
HPO
and HCl:
2
4
2−
[HPO
]
4
7.00 = 7.20 + log
[H
PO
]
2
4
2−
2−
2−
[HPO
]
molesHPO
molesHPO
4
4
4
= 0.63096 =
=
−
−
0.100
[H
PO
]
molesH
PO
2
4
2
4
2−
molesHPO
= 0.063096
4
2-
+
-
HPO
+
H
H
PO
+ H
O
→
4
2
4
2
Start:
y
0.100
0
React:
-0.100
-0.100
+0.100
€
End:
0.063096
0
0.100
-
y = 0.163096 moles HPO
4
HCl: 0.100 moles x L/1.00 moles x 1000mL/1L = 100. mL 1.00 M HCl
:
Na
HPO
0.163096 moles x 141.96 g/mol = 23 g Na
HPO
2
4
2
4
Plus water to give 1 L total volume
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