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.
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
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0.01 M NaOH,
M = 1 * 10-2
pOH = 2
pH = 2
Kp = Kc (RT)Dng
36 * 10–2 = Kc (0.0821 * 300)–1
Kc = 0.36 * 0.0821 * 300 = 8.86 » 9
A(g) ->B(g) + (g)
Initial moles n 0 &nbs
On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.
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