The ionization constant of nitrous acidis 4.5x 10^4 calculate the ph
of 0.0LM solution of nitric acid in water
Н
[Hint: 4 NO, +420 = H,O+NO3 Raca]
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Answer:
Since NaNO
2
is the salt of a strong base (NaOH) and a weak acid (HNO
2
) and NO
2
−
+H
2
O↔HNO
2
+OH
−
So,
K
b
=
[NO
2
−
]
[HNO
2
][OH
−
]
⇒
K
a
K
w
=
4.5×10
−14
10
−14
=.22×10
−10
Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:
[NO
2
−
]=.04−x;0.04
[HNO
2
]=x
[OH
−
]=x
K
b
=
0.04
x
2
=0.22×10
−10
x
2
=.0088×10
−10
x=0.093×10
−5
∴[OH
−
]=0.093×10
−5
M
[H
3
O
+
]=
0.093×10
−5
10
−14
=10.75×10
−9
M
⇒pH=−log(10.75×10
−9
)
=7.96
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