calculate the normality of 7.3gm hcl is present in 250ml of solution
Answers
Answer:
Explanation:
Number of moles of HCl=
36.5
7.3g
=0.2 moles
Thus molarity of the solutions=
v
n
=
20
0.2
M
HCl
=0.01M=10
−2
M
Now, 50ml of this solution is taken and diluted to 250ml. Thus, new concentration of solution will be
⇒M
1
V
1
=M
2
V
2
⇒0.01M×50ml=M
2
×250ml
M
2
=2×10
−3
M
This is the concentration of diluted HCl solution as 40ml of this solution dilute 20ml of Ba(OH)
2
Thus, concentration of Ba(OH)
2
is
⇒40×2×10
−3
=20×M
3
×2 nf=2 for Ba(OH)
2
⇒M
3
=2×10
−3
M
Thus, this is the concentration of Ba(OH)
2
s
Ba
(OH)
2
⇌
s
Ba
2+
+
2s
2
O
−
H
Thus concentration of
O
−
H is
⇒[
−
OH]=2×(2×10
−3
M)
⇒[OH
−
]=4×10
−3
M
pOH=−log[OH]=2.397
as pH=14−pOH
pH=14−2.397
pH=11.6
Thus, pH of Ba(OH)
2
is 11.6 .
Answer:
Number of moles of HCl=
36.5
7.3g
=0.2 moles
Thus molarity of the solutions=
v
n
=
20
0.2
M
HCl
=0.01M=10
−2
M
Now, 50ml of this solution is taken and diluted to 250ml. Thus, new concentration of solution will be
⇒M
1
V
1
=M
2
V
2
⇒0.01M×50ml=M
2
×250ml
M
2
=2×10
−3
M
This is the concentration of diluted HCl solution as 40ml of this solution dilute 20ml of Ba(OH)
2
Thus, concentration of Ba(OH)
2
is
⇒40×2×10
−3
=20×M
3
×2 nf=2 for Ba(OH)
2
⇒M
3
=2×10
−3
M
Thus, this is the concentration of Ba(OH)
2
s
Ba
(OH)
2
⇌
s
Ba
2+
+
2s
2
O
−
H
Thus concentration of
O
−
H is
⇒[
−
OH]=2×(2×10
−3
M)
⇒[OH
−
]=4×10
−3
M
pOH=−log[OH]=2.397
as pH=14−pOH
pH=14−2.397
pH=11.6
Thus, pH of Ba(OH)
2
is 11.6