what is the pH of the solution containing 1.95 gram of H2 S o4 mole per dm3
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Answered by
0
Answer:
ExH
=
1.40
...
...
.
.
Explanation:
We assume complete dissociation of the sulfuric acid:
H 2 S O 4 ( a q ) +2 H2 O ( l) → 2
S
O
2
−
4
And so
[
H
2
S
O
4
]
≡
1.95
⋅
g
98.08
⋅
g
⋅
m
o
l
−
1
1
⋅
d
m
3
=
0.0199
⋅
m
o
l
⋅
L
−
1
. However, this is a FORMAL concentration and corresponds to
[
S
O
2
−
4
]
.
[
H
3
O
+
]
=
0.0398
⋅
m
o
l
⋅
L
−
1
, i.e. TWICE the formal concentration.
Now since,
p
H
=
−
log
10
[
H
3
O
+
]
,
p
H
=
−
log
10
(
0.0398
)
=
?
?
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