7.61. The ionization constant of nitrous acid is 4.5×10−4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
7.61. The ionization constant of nitrous acid is 4.5×10−4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
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1 Answer
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Since NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2);
NO2− + H2O ↔ HNO2 + OH−
Then, Kb= [HNO2] [OH−] / [NO2−]
⇒ Kw/ Ka = (10−14) / (4.5×10−4)
= 0.22×10−10
Now, if x moles of the salt undergo hydrolysis, and then the concentration of various species present in the solution will be:
[NO2−]= 0.04−x; 0.04
[HNO2]= x
[OH−]= x
Kb= x2 / 0.04
= 0.22×10−10
x2= 0.0088×10−10
x= 0.093×10−5
∴ [OH−]= 0.093×10−5 M
[H3O+]=10−14 / 0.093×10−5=10.75×10−9 M
⇒ pH=
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