How would you prepare 1L of a .050 M phosphate buffer at pH 7.5 using crystalline K2HPO4 and a solution of 1 M HCl? pka of H2PO4 = 7.2

Answer 1

You want the pH of the buffer to be 7.5; therefore, calculate the ratio of base to acid using the Henderson-Hasselbalch equation.
pH = 7.2 + log(base/acid).
Solve for B/A. I get something like 2.

You have two unknowns; you need another equation. That is acid + base = 0.05
Using these two equation, solve for acid and base, convert base to grams using the molar mass of K2HPO4 and acid to mL using the molarity of HCl.
Post your work if you get stuck.

Answer 2

I got the first part when I have to find the ratio for base to acid. I didn't know where to go from there. I tried using the C1V1=C2V2 equation but again I am lost.

Answer 3

1.Weigh out 8.7 g k2hpo4(0.05 mole ×174 g/mole=8.7g).

2.dissolve it in small amount of water.

3.since the base/ acid ratio =2/1 you must convert 1third of the base (k2hpo4 )to the acid form - so you must add 0.05/3 L of 1M hcl to the solution.

4.dilute the resulting mix to 1 L .

bon appé :-)

Answer 4

We need to find the amount of crystalline powder we need to weigh
in order to have 0.05 M of buffer
and the
volume of acid (1M HCl) we need to add and then make up the
volume
to 1L with water
...
.then we have our
buffer at pH 7.5.
N
umber
of moles = m/M
n = m/M
0.05mol/l
= m/174
g/mol
m = 8.7g
So we have to weigh 8.7g of K
2
HPO
4
and dissolve it in a
small
amount of water
How much of 1 M HCL do we need?
HCL

H
+
+ Cl
-
and the H
+
will go bind to our buffe
r
K
2
HPO
4
ioni
z
es in water to give 2K
+
and HPO
4
2
-
Then the
H
+
from HCL will bind the HPO
4
2
-
to give H
2
PO4
-
HPO
4
2
-
+
H
+
-
>
H
2
PO
4
-
If we know how much of acid
(
H
2
PO
4
-
)
is required
then we will know how
much of HCL we need to add since
for every 1mol of
acid
we need 1 mol of HCL (
1 mol of
H
+
from HCL)
Without considering pH we can easily calculate the volume we need
to make 0.05M of buffer
:
C
1
= 1M
C
2
= 0.05M
V
1
=
?
V
2
= 1
L
C
1
V
1
=
C
2
V
2
1M x V
1
= 0.05
M
x 1L
V
1
= 0.05L
This is
how much we need to make 0.05M of buffer at any pH b
ut
we don
’
t know how much of this
we
need
to adjust the pH to 7.5. So we need to use the Henderson
-
Hasselbalch equation
to find out how
much of the acid we need
pH =
pKa + log [base]
/[acid]
7.5 = 7.2 + log [base]/
[
acid]
0.3 = log [base]/[acid]
10
0.3
=
[base]/[acid]
2 = [base]/[acid]
2/
1
= base/acid
S
o
1/3 of
volume of acid
is
re
quired
to make pH 7.5
Volume of HCL required =
1/3 x
V
1
= 1/3 x
0.05L
= 0.0167L
= 16.67ml
(multiply by 1000 since 1000ml = 1L)
So to prepare your buffer of 0.05M at pH 7.5 you weigh out
8.7 g of
K
2
HPO
4
and dissolve it in a small amount
of water and add 16.67ml of 1M HCL and then make t
he volume up to 1L with water
...
.t
hen you have your